One Sided Limits, Graphs, Continuity, Infinity, Absolute Value, Squeeze Thereom - Calculus Review

This video is about

limits

, we will focus on finding

limits

graphically, analytically and even finding one-

sided

limits. Boundaries in Infinity, we're going to talk about

continuity

and some other things too, so let's get started, say if we want. evaluate the limit as approaches -3 from the left side and we are looking for f ofx which is associated with variable Y so when X approaches -3 from the left side the

value

of Y is approximately two so that is the answer for this now, what is the limit when X gets closer? -3 but from the right side, so what is the

value

of Y when X approaches ne3 from the right?

So if we follow the function, notice that the value of Y is approximately three. Now what is the limit when X approaches -3 from either side from the left? side and the right side, these are known as one-

sided

limits because they do not coincide with the limit when , what is the The value of Y when Now, what kind of dis

continuity

do we have? AT3 is is a point discontinuity, a jump discontinuity or an infinite discontinuity, this is known as a jump discontinuity, you can see the jump or disconnection between the two

graphs

. Now a jump discontinuity, is it removable or non-removable? a jump discontinuity is a non-removable discontinuity, now let's move on to the next, what is the limit when X approaches -1 from the left side, so negative one is on the vertical ASM toote?

So if we follow the curve from the left side, notice that it goes up towards Infinity. Now what about the limit? when X approaches -1 but from the right side, if we follow the function from the right side, it also goes up to

infinity

. What happens to the limit when X approaches ne1 from either side because the left side and the right side are equal? It's going to be Positive Infinity, now you have to be careful with this one because we have a vertical ASM toote at Nega one, so the graph shows unlimited behavior at this point and some textbooks will say that the limit does not exist and its justification behind of that.

If

infinity

is not actually a number, the graph keeps going up, it does not converge to a specific number and that is why some textbooks may say that the limit does not exist for this example and others will simply write infinity, but keep in mind that infinity is no, it doesn't converge to a specific number, it just keeps going up forever. What about F of 1? Which is the answer? Notice that there is no closed circle along the ASM vertical totes, so there is no value of y when X is 1, so this function does not exist or this value does not exist at What type of discontinuity do we have at x = 1 since we have a vertical ASM?

This is an infinite or infinite discontinuity. Infinite discontinuities cannot be eliminated. What is the limit? as X approaches one from the left side, here is the value of one on the x axis as we approach it from the left side, the value of Y is four. Now, what is the limit as X approaches a positive one, but from the right side, as we approach it? from the right side, the value of Y is still four, so what is the limit when X approaches one from any side because the left side and the right side are equal?

The limit exists, it is equal to four now, what is f of one? So when X is exactly one, what is the value of y? ​​So in this case, we need to look for the circle Clos and it has a value of one, so F of 1 is equal to 1. Now, because the limit is not equal to the function at xal 1, we have a discontinuity now, what kind of discontinuity do we have? This is known as a point discontinuity, also a removable discontinuity. What is the limit when X approaches three from the left side? What is the answer for this one?

Also try these too. What is the limit? X approaches three from the right side and when X approaches three from either side and what is f of three, once again we have a vertical ASM toote from the left side. Notice that it goes up to Infinity, but from the right side it goes down to Negative Infinity, so because these two don't match, the limit doesn't exist and F of three is not defined and the last example I gave doesn't exist, but undefined is a better answer because let's say if f ofx is equal to 1x where we have a vertical ASM toote at x is equal to z if you enter zero this is set to undefined so it is not defined in a vertical ASM toote what is the limit when X approaches 2 from the left side and further what is the limit when X approaches 2 from the right side and when What is the limit when X approaches two from the left side?

So if we follow the curve from the left side as we approach two, notice that the value of Y is two now, as X approaches two from the right side, the value of Y still converges to two, so Therefore, the limit when X approaches two from any side is two, since the left side and the right side are equal now, what is F of? two, so when X is exactly two, what is the value of Y? So we need to look at the near circle and when X is two we can see that Y is two.

Note that all the values ​​agree with each other, so xal 2 is continuous. the function is defined, the limit exists and the limit is equal to the function Now sometimes in a proof you may be asked to prove that a limit exists at a certain point, which is kind of like what we did in the last example, so there is something called a three-step continuity test, the first thing you do is make sure that the function is defined at a certain point or at the point of interest, in this case at xal a, and second is to make sure that the limit exists at xal a or when that F of two is equal to two in the last example that we did it like this, we define F of a, step two, we need to prove that the limit exists, so we need to check the limit of the left side and the right side to that the limit as X approaches two from the left for f ofx would be equal to two and the limit as a two on either side exists and is equal to two now the third step is to make the statement that the limit as which you must follow whenever you want to prove that a certain point is continuous, so that's the three-step continuity test. so now let's work on some examples where we can evaluate a limit algebraically or analytically what is the limit as X approaches 2 for the function 20/3x + 4 so how can we evaluate this limit now?

The first thing you want to do or What I want to check is to see if you can substitute two for x, and for this example, direct substitution works as long as you don't get a zero in the denominator of the fraction, you can substitute two, so 3 * 2 + 4 3 * 2 is 6 and 6 + 4 is 10 so we can cancel a zero so 20 10 is the same as 2 over 1 which is two and this is how you can evaluate the limit of a function, let's try another example like that, feel free to pause the video as you work through this example, what is the limit as 3 is 3 + 5 - 8 3 + 5 is 8 8 - 8 is 0, so that's the answer for this particular problem.

Now what about this one? What is the limit as X approaches 5 for this function? x - 5 / x^2 - 25. Now, what will happen if we try to use direct? substitution, so if we enter five it will be 5 minus 5 ID 5^ 2 - 25 now 5 - 5 is 0 5 square that is 25 so 25 minus 25 is z 0 over Z is indeterminant, you don't know if it's going to be zero if it's going to be infinite if it's going to be 8 we don't know what this is because we get a zero at the bottom we can't use direct substitution we need to use another method in a situation like this what?

What you want to do is see if you can factor the expression, so how can we factor x^2 minus 25? What we have is the difference of squares, if you have a 2 - b^ 2 to factor, it will be a + b * a minus B the square root of x s is X and the OT square of 25 is 5 we are going to have a plus and a less, so notice that we can cancel x - 5, so once x - 5 at the bottom disappears now we can evaluate the limit using direct substitution, so let's substitute x for 5 so that it is 1 over 5 + 5 and 5 + 5 is 10, so it's 1 over 10 1 over 10 as a decimal is 0.1, so this is the answer prove this is the limit as X approaches 4 for x^2 - 4x / x^2 + 3x - 28 so should we factor or can we use direct substitution?

Well, let's see if the direct replacement is going to work, so let's connect four at the bottom. is 4 2 + 3 * 4 - 28 4 2 is 16 3 * 4 is 12 16 and 12 is 28 so we get a zero in the denominator which will not help us to get the answer, so we need to factor, so how can we factor? x^2 - 4x all we can do here is just remove the GCD, we can get an How can we factor a trinomial with a leading coefficient of one? For this example find two numbers that multiply to -28 but add three, so the factors of 28 are 1 and -28 2 and 4 4 and -7 or -4 and 74 + 7 is pos3 and multiply to -28 , so factoring it out is it's going to be x - 4 * x + 7 notice that we can cancel out and x - 4 so what we are left with is the limit as the bottom part so now we can use direct substitution so it is four / 4 + 7 4 + 7 is 11 so the final answer is 4 over 11 consider this problem what is the limit when X approaches Z for this function 3 + x ^2 - 9 / X so for this example I can't use direct substitution otherwise we will get a zero in the denominator.

Now we can frustrate 3 + x^2 or we can factor it out, but let's go ahead and frustrate 3 + x^2, so this is the same as the limit. as for each step you need to rewrite the word limit until you substitute zero into X. Some teachers will take points from you if you don't rewrite the limit expression. Now notice that we can cancel a nine, so now we have X2 and we can combine. 3x + 3x which is 6X so now let's factor an cancel this x, so once this x is gone, you can now substitute zero into You might get stuck in terms of what to do to simplify a limit before you can use direct substitution.

You can always use your calculator to get the answer. This is very useful if, first of all, you can use a calculator on a test and if it's multiple choice, so let's choose. a number that is very close to zero, so let's figure out what F of 0.1 is equal to, if you join this into 3 +. 1^2 - 9 / .1 will be 3.12 - 9 takes that result which is 61 / .1 and is 6.1. Now keep in mind that our answer was six. Now notice what happens if we choose a value closer to zero, so use 01 instead of 0.1 so 3.01 2 - 9 that's 0601 / 01 this is equal to 6 6.01 notice that the answer is getting closer to Six, it's very close to it so you can see it's going to converge to six which means the limit exists now if you try a negative value let's say negative 001 then it's going to be 3 minus .1 squared - 9/1 3 - 01 is 2999 to the power the power 3 or excuse me raised to the second power - 9/1 you should get 5999, which is approximately six, as long as you enter a number that is very close to zero but not exactly zero, you can get the answer by direct substitution.

It works, try this, what is the limit as X approaches -3 for function x^? 2 - 9 / 2x^2 + 7 x + three, so whenever you have a rational function where you have a polynomial divided by a polynomial, your initial instinct is to factor and cancel, so that on top of that we have the difference of perfect squares, like this which uses an equation a s minus b^2 is A + B A minus B to factor x^2 - 9 will be x + 3 * x - 3 but whatever you do, don't forget to rewrite the limit expression, otherwise you may lose points your tests now, how can we factor a trinomial where the leading coefficient is not one?

So how can we do that? Let's multiply 2 andthree. 2 * 3 is 6 and we need two numbers that multiply up to six but add up to seven, which is the middle term. and this is 1 and six 1 * 6 is 6 but 1 + 6 is 7 so I'm going to factor this expression aside and then I'm going to put the answer back where it belongs now what you want to do is what you want to replace 7x with 6X + 1 terms, the greatest common factor is 2x 2x^2 / 2x is one, if you factor one, it will simply be x + 3. Now, if these two factors are equal, then you are on the right track, you can factor x + 3 and what goes in the second parenthesis is the material. that's outside that's uh 2x + 1 and then we can put that in here so now we can see that we can cancel out X+ 3 once we do that we can use direct substitution to get the answer so that the limit when for the function x - 3 2x + 1 is going to be -3 - 3 over 2 * -3 + 1 -3 - 3 es6 2 * -3es -6 and -6 + 1 es 5 whenever you have two negative numbers divided by each other , will be a positive result, so the answer is POS 6 out of 5, which is 1.2 as a decimal.

Here is another example you can try, what is it? limit when Can we do it so what we have is the difference of perfect cubes and the equation is a minus B * a^ 2 + a + b^2 so a the 3r is X cub and 125 is B the 3r to find a and b? We need to take the cube root of x Cube, which is is 5 * 5, which is 25, so we can also write our expression. Let's make some space first, so what we now have is the limit as X approaches POS 5 and at the bottom we have x - 5 * x^2 + 5x + 25. so we can cancel we can replace x with five so it will be 1/5^ 2 + 5 * 5 + 25 5^ 2 is 25 5 * 5 is also 25 plus another 25 which is 25 * 3 which is 1 over 75 now what is the limit when X approaches zero for this function 2 + x to the power 3 - 8 / and get the answer this way, but I'm going to use the equation, so what we need to keep in mind is that a is 2 + x and B is 8, so a if a the 3r is 2 + x 3r a is 2 + x and if b^ the 3 is 8 then B is the cube root of 8, which is 2, so we have 2 + x - 2 a^ 2, that will be 2 + x squared and then plus a, so B is 2 2 and a is 2 + x, then plus b^ 2, which is 2^ 2, so that's simply 4, that's how we can factor it out, but notice that the two cancel, so what we now have is the limit as X approaches to zero at the top, we are left with one that means we can now plug zero into the equation, so now what we have is 2 + 0 2, that's this part + 4 + 2x, which is just 2 * 0 and then plus these other four, so 2 + 0 is simply 2 and 2^ 2 is 4 plus another four plus another 4 4 + 4 + 4 is the same as 4 * 3, which is 12, multiplication is simply repeated addition, so the final answer is 12.

Now we're going to confirm it using direct substitution, so pick a number that's very close to zero that we can plug in and see if it's going to give us 12, so let's figure out what F is 0.01, so it's going to be 2 + 01, which is 2.01 ra 3r - 88 over 01 2.01 to the power of 3 is approximately 8.1 12061 minus 8 which is 1 12061 ID 01 and the answer you get is 12.06 01. Now, to be safe, let's choose a smaller number, so let's try 0.001, at least You should do it twice with two different numbers because it does change. so it doesn't converge, but if it's still 12, then you know you have the answer, so it's going to be 2.1 to the power of 3r - 8 / .1 2.1 to the power of 3 is 8.12 minus 8 and then divide that by 01 so the answer is about 12.6 , so notice that this is even closer to 12, so the answer is 12.

Now let's say if you have a question or a problem like this, how would you evaluate this limit? What would you do to be able to use direct substitution? To work, you can't substitute 9 directly because the square of 9 is 3 3 - 3 is 0, but you can enter a number very close to N, so let's figure out what the value of 8.9 is, so 9 - 8.9 / 3 - < TK 8.9 9 - 8.9 is 0.1 and 3 minus the < TK of 8.9 is 06713 something like that, so 0.1 divided by that number gives approximately 5.98, so it rounds to six, so if we try 8.99, that's 9 - 8.99 over 3 - there is < TK of 8.99 9 - 8.99 is 01 3 - the square root of 8.99 will be 0.166 7 13, so if you divide those two numbers you should get 5998, which is even closer of six, so we can see that the final answer should be six but how can we get the answer without using a calculator?

If you see a radical, you want to multiply the top and bottom by the conjugate of the expression containing the radical, so the conjugate of 3 - radical X is 3 + radical X. You simply have to change one sign from negative to positive and everything you do downwards you have to do upwards now for one of these you want to frustrate but for the other you don't wherever you see the conjugate that is the part you want to frustrate because those two differ only by the sign one is plus the another is minus you just want to frustrate the 3-rootx and the 3 plus rootx because the two middle terms will cancel these two expressions that are not conjugated with each other do not frustrate it leave it in its factored form if you frustrate it life will be difficult and it will be much more hard to solve now let's go ahead and frustrate the denominator so 3 * 3 is 9 3 * < tkx is 3un the square roots cancel and you get what is inside which is inside now up, we'll just leave it as it is so we can cancel these two terms and now what we do What we have left over is this particular expression, at this point we can cancel the 9 - x, so now we have the limit when approaches 9 for this function, so we can use substitution at this point, so it will be 3 + the square < TK of 9 and the square < TK of 9 is 3, so 3 + 3 is 6, which is the answer we got with the calculator, so when you see a radical multiply the top and bottom by the conjugant, so here's an example you can try yourself. and see if you can get the answer, so what is the limit as X approaches Z for 4 + x or radical 4 + x - < TK4/X?

So, as before, we are going to multiply the top and the bottom by the conjugate which is the same but instead of a minus sign it will be a plus sign and whatever you do on the top of the fraction you must do the same on the bottom of the fraction, so remember to only frustrate the conjugates. Don't frustrate these two. or distribute the will cancel out, but I will write it if you want to see it so that radical 4 + x * < TK4 is simply < TK4 root4 + x and these two will be the same but with a negative sign TK4 * < TK4 is simply -4 the radicals will disappear and in the part below we will leave it in its factored form so that we can now cancel.

Note that these two terms cancel 4 +4 is zero, so what we are left with now is this expression. Note that we can cancel the radical which we have < TK4 + a 4. the square root of 4 is 2 and 2 + 2 is 4, so the final answer is 1 over 4, so that's it for that problem. What if you have a problem like this? What should we do to evaluate the limit of this function? Whenever you have a complex fraction, what you should do is multiply the top and bottom, not by the conjugate but by the common denominator. You want to multiply the top and bottom by something that eliminates the fractions, so look at the denominator of those two. fractions we don't have fractions in the denominator, so we'll leave it in its factored form, so 3x * 1X what cancels this is the same as 1X * 3x over 1, the X variables will cancel and there will be three, so now it will be three , how about 1/3 * 3x?

This time all three will cancel out and you will simply get X, but there is a negative sign in front of it at the bottom, leave it in its factored form now. Can we cancel 3 - x and x - 3? They look very similar but right now we can't cancel it out yet, so what we have to do is take a NE 1 from the numerator and reverse the terms if we factor Nega 1 thex. will change to positive and 3 * 3 is 9 so the final answer is -1 over 9 try this so here we have another complex fraction so we need to get rid of the 4 and the 4 plus X so in this example the common denominator is 4 * 4 + x and everything Whatever you do at the top you must also do at the bottom, so if we take 1 over 4 + x and multiply by 4 * 4 + x, which cancels out and what remains, you will notice that the 4 plus what the four will remain so now if we multiply this by 1/4, what cancels out and what remains, notice that the four will cancel out and the 4 + x will remain, but don't forget that there is a negative sign in front of it, so which will be negative and then the 4 + x and at the bottom there were no fractions to remove so let's just rewrite it like this four and 4 + x , so it will be 1 * 4 and 4 + 0 4 + 0 is simply 4, so we have 4 * 4 at the bottom, which is 16, so the final answer is -1 plus. 16 so that's what you have to do for that problem now, what would happen if you saw a question like this?

It is a little longer and more difficult, but it can be solved. Go ahead and find the limit for this expression as X approaches zero. Note that we have two radicals and we have a complex fraction, so should we multiply by the conjugate or by the common denominator? The answer is both personally. I prefer to eliminate the fractions before multiplying by the conjugate and that is what I am going to do. Do it, I'm going to multiply the top and the bottom by the radical x + 5 * root 5, so if we take this term and distribute it to this, notice that the rootx + 5 will cancel out and the five will remain, so It's going to be just five in the top root five specifically now, if we take this and multiply it by 1 over a 5a 5 will cancel out and we'll be left with the radical x + 5 and at the bottom we'll just leave it as is.

It's like this now that we have cleaned up the fractions now is a good time to multiply the top and bottom by the conjugate of the numerator so it will be 5 plus otx + 5 and at the bottom we are going to do the same but No we're going to foil it at the bottom nor distribute it, so < TK 5 * TK 5 is just five, this term will cancel out with this term, so you can write it down if you want, but you know it's going to cancel out and then finally these two terms will become ax + 5 rootx + 5 * root otx + 5 you have to be careful with every step in a problem like this because if you make a mistake that's it, the whole question is ruined so now let's distribute the negative 1 so that it is 5 - x - 5 divided by everything which is at the bottom so we can cancel five plus 5 and negate five zero sum now what else can we cancel at this point?

See what we can cancel? Notice that we can cancel anso we get rid of the 0 + 5 so 0 + 5 is simply five so we have otk 5 * otk 5 and in parentheses TK 5 plus T 5 so what is the square of 5 times 5 times < TK of 5 if we multiply? these two together will be the square root of 25 5 * 5 is 25 now if you add the root of 5 + a 5 it is like adding 1 + 1 1 + 1 is 2 so you have 2a 5 the OT square of 25 is five so the final answer will be well, we need to multiply five and two, so that's 10, so that's 1/10a 5, but maybe that's not the final answer because we could rationalize the denominator, so let's multiply up and down by root five, like this which above is root of five the square root of 5 * the square Ro T of five we know it is five so the final answer and its completely simplified form is theare < TK of 5/50 this is now, let's

review

some basic properties of limits , so let's say that if the limit as X approaches C for the function f ofx is equal to 4 and the limit as the limit as X approaches C for this expression f ofx plus?

G of X, what is the answer, all you need to do is simply add f and g, the value of f as so if you see a question like that, that's pretty much all you have to do to get the answer. What's wrong with this expression? The limit as X approaches c f ofx times G ofx, so this expression is equal to the limit as f value of x be four and for G be 5 so it is 4 * 5 which will give you a value of 20. Let's try another limit as is the answer so this is the same as the limit as that's the answer for that one, let's try one more example, so this time the expression will be 3 F ofx over G ofx minus F ofx, so f ofx is equal to 4 as X approaches C and for G it's five , so it will be 5 - 4 3 * 4 is equal to 12 5 - 4 is 1 and 12 / 1 is 12, so this is the final answer for this example.

Consider the following piecewise function, so f ofx is equal to 2x + 1 when is greater than 4 and is also equal to 7 when thing, the only thing we have to keep in mind is -3 and 4, it can be continuous or it can not be in those two points as long as If you have a fraction, the point of discontinuity usually occurs when the denominator is equal to zero. If you have a radical, just remember that you can't have any negative numbers inside a square root if you have an even index number and for records. There can never be a zero or a negative number within a record, but now we have none of that in this particular case.

If you want to quickly determine the discontinuity points, enter -3 in these two expressions if they give you the same. The y value will then be continuous in g3, so 2 * -3 + 1 is -6 + 1, which is 5 now for the second expression, if we insert -3 into X, it will be -32 es 9 9 - 14 es5, so because those two expressions have the same value of y at x = -3, it will be continuous at that point now if we connect four in these two expressions, let's see if we get the same value of y, then 4^ 2 -4 and 3 * 4 - 4 4 2 is 16 16 - 14 is 2 3 * 4 is 12 12 - 4 is 8 since these two are not equal, it is discontinuous in xal 4, this is how you can quickly tell if it will be continuous or discontinuous in a circular function Now, sometimes you may have to justify your work, so how can we use three-step continuity tests to prove that it is continuous at -3 but discontinuous at 4 and also determine the type of discontinuity at xal 4?

I'm sure the function is defined at -3, so what is f of 3? Should we use this part of the function or the second part to find it? Notice that the underline is in the second part of the equation, so we have to substitute a -3 into this equation because that's when it can be equal to -3 so it's -3 2 - 14 that's 9 - 14 that's 5 so f of -3 is defined now what is the limit when X approaches -3 from the left side to the left side What function should we use? If we use 2x + 1 or x^2 - 14, the left side of 3 3 would be a number like -3.1 if you draw a number line, here is -3 -3.1 is to the left of 3 and -2.9 . is to the right now -3.1 is less than 3 so we have to use the first part of the function 2x + 1 so we are going to plug -3 into that equation and that will give us 5 now we also need to find the limit as approaches -3 from the right side, we have to use this function and it will be -32 - 14, which is 9 - 14, that is ne5 too, because the left side and the right side limit because they are the same way, the limit as X approaches -3 exists and is equal to 5, so now we can state that the limit as It can be shown that it is continuous at x = -3 now let's show that it is discontinuous at x = 4 so first let's see if F of four is defined what is the value of f of four then when X is exactly four what is the value of y so X is four at this point, so Y is seven, so the function is defined at xal 4.

Now, what is the limit as So a number to the left of four is like 3.9, which is less than four, so it's within this range between -3 and 4, so we have to use this part of the circle function, so it will be 4 ^2 - 14 which is 16 - 14 that is 2. Now what is the limit when X approaches 4 from the right side so a number is greater than 4 we need to use this equation 3x - 4 so that is 3 * 4 - 4 that is 12 - 4 and that is 8 because the limit of the left side and the right side do not coincide, the limit does not exist, therefore it is discontinuous at x = 4 now what type of discontinuity have?

If we graph towards the left side we have a value of two and it is an open circle and on the right side we have a value of eight, so we have an open circle. at eight and we have a closed circle at 4 point 7 that should be somewhere in this region now this function is a parabola so it's probably something that looks like this now the right side 3x - 4 is a linear equation with a slope of three then it should have a shape similar to that, but notice that we have a jump discontinuity. The reason why it is a jump discontinuity is because the left side limit and the right side limit do not coincide, one was two and the other is eight, whenever you see that it is there is going to be a jump discontinuity now consider this another circular function 7 x + 1 and CX x^2 + 3 and for this X is less than or equal to 2 and for the second part value of C that makes the circular expression or function continuous at x = 2.

So what can we do to find that value of C? Now to make it continuous they must have the same value of y 7x + 1 must be equal to C x^2 + 3 when 7 * 2 + 1 is equal to C * 2^ 2 + 3 7 * 2 is 4 2^ 2 is 4 and 14 + 1 is 15 so we have 15 is equal to 4 C+ 3 so let's subtract three from both sides 15 - 3 is 12 and then divide both sides by four so 12/4 is 3 so C is equal to 3 at that point the function will be continuous when C has a value of three let's try one more example so let's say that if we have ax + 7 2x + B and ax2 minus 32 so for the first part 1 and x = 4 feel free to pause the video while you solve this example, so we are going to have a system of equations that we need to set the first two equal to each other when X is equal to one, so x + 7 will be equal to 2x + B when Let's subtract B from both sides and subtract 7 from both sides so that these two cancel.

On the left side it will be minus B and on the right side 2 - 7 is 5, so we get the expression a - B. is equal to 5 now we are going to equate the last two together when X is 4 then 2x + B is equal ax^2 - 32 when we subtract B from both sides, so on the right side we have 16 a minus B and on the left side it's 40, so now let's solve. the system of equations that uses elimination you can use substitution if you want, but I chose to use elimination, so 16 a minus B equals 40 and a - b equals 5 when we add the two equations, we need to add them such that b will cancel o a but it is easier to cancel B, let's multiply the second equation by 1, so the first equation will remain the same but the second equation will be negative a plus b, which is equal to POS 5 if we multiply by minus 1 the reason why we did that is so that when we add the two equations B we will cancel minus B plus b is zero so 16 a + a is 15 a 40 + 5 is 45 so if we divide both sides by 15 we could see that a is equal to 3 now that we know that we can solve for b let's use this equation so that -3 + B is equal to 5 then let's add three to both sides so that 5 + 3 is 8 so that B is equal to 8 then, when a is 3 and b is 8, the function will be continuous at x = 1 and at xal 4 now let's

review

the largest integer function, the largest integer of Here's a quick and easy way to find the answer.

Let's put two 3's and one on a number line. Now, where is 2.4 on the number line? 2.4 is somewhere in this region, so you want to choose. the integer to the left of 2.4 is two, so the integer greater than 2.4 is two, you must choose an integer that is less than or equal to 2.4. Now there are other numbers that are less than two. 01 -2 these are integers too, but the largest of them is two, that's why the answer is two, so now let's try with more examples of what is the largest energy of 3.9, 1.6 and 3.1, so go ahead and try these, let's focus on 3.9, so let's write three as a center 2 and four, so 3.9 is very close to four, but to find the largest integer that is less than 3.9, choose the integer on the left, so the answer is three.

Now what about 1.6, so we graph 1 0 and -2 you always want? choose a number on the left and right now where is 1.6. 1.6 is in this region, so we need to choose the integer that is to the left of 1.6, so -2 now for the last one we are going to plot -3 -4 and -2 -3.1 is to the left of -3, so the largest integer of -3.1 chooses the integer on the left which is 4 and that is the answer, so now let's apply the largest integer function to limit what is the limit when X approaches 2 . for the function 3 minus the largest integer function of X, so what is the answer?

Now what you need to do is find the limits of the left and right sides, so let's first find the left side of the limit, what is the limit when X approaches 2? From the left side, let's say if we enter a number that is less than two, like 1.9, that is to the left of two, so 3 minus the largest integer of 1.9, what is the answer? So if we plot 2 3 and 1 1.9 is here, then any number to the left of two, but very close to two, will be in this region and we have to choose the integer that is to the left, so it is one, so the largest integer of 1.9 is 1, so it will be 3 minus 1. which is 2, so that's the value for the left side limit.

Now let's evaluate the right side of the limit, so what is the limit as X approaches two from the right side? So this time we need to enter a number that is very close to two but larger. that two or from the right of two then we can prove 2.1 so if we do the number line here is two this is one this is three two to the right is in this region so what is the largest integer of 2.1 so choose the integer to the left of 2.1, which is two, so it will be 3 - 2, which is 1 now because the limits of the left and right side because they are not equal, the limit does not exist, they do not coincide with the side of the limits that exist, but the limit itself when X approaches two from both sides does not exist.

Try this, the limit as X approaches 3. Well, actually -3 for the expression 4 plus thewith this one, then what is the limit when X approaches Pi/2 for Tan x? Which is the answer? So Pi/2 is about 90°, let's look at the left side and the right side, so to the left of 90 it's like 89.9, so if you enter tan 89.9 into the calculator, make sure it's in mode of degrees, you should get about 573 now, if you write tan 89.99, this will be about 5.730 to notice that as X approaches 90° from the left or the tangent Pi/2 approaches positive Infinity, what happens to the limit when X approaches 90 or Pi/2 from the right side, so choose a number that is greater than 90 but close to it? so if we plug Tangent 90.1 you should get 5 573 and if you write tan 9.01 you should get - 5730, therefore the limit when X approaches 90 or Pi/2 from the right side is negative infinity and from the limits of the sides left and right do not match the limit does not exist now let's get the answer using a graph the graph of the tangent has a vertical ASM toote at piun / 2 and Pi / 2 the period is pi so the next vertical ASM toote is at 3 pi / 2 3 pi/ 2 and pi/ 2 differ in pi and the tangent is an increase in the function, so it looks like this and it will repeat, so this pattern will continue to repeat, but we will focus on pi over two as Let us get closer to you. / 2 from the left side notice it goes to Positive Infinity so from the left side we plugged in 89.9 and we got Positive Infinity as we get closer to pi/ 2 from the right side it becomes negative infinity and that's why the limit does not exist. now you can also get the signs using a unit circle Pi/2 or 90 is in positive and AIS so as we approach 90 from the left side i.e. 89.9 we notice that we are in quadrant 1, the tangent is positive in quadrant one, so it is positive Infinity when we plug 89.9 to the right of 90, which is like 90.1. 90.1 is in quadrant 2 and in quadrant 2 the tangent is negative, so one side of Pi/2 is negative, the other is positive, therefore the limit does not exist and what about this, the limit when approaches zero for the secant function what is the secant answer is 1/cosine and notice we can solve it using direct substitution what is cosine 0 now if you write cosine 0 you should get one 1/ times one? is one so that is the limit for this particular problem now what is the limit as if you enter exactly zero it will not be defined, but if we enter a number close to zero maybe we can get the answer, so let's try 90 from the left side, now we know it will be positive infinity or negative infinity because we cannot cancel the cosine , I just got to know what the sign is to the left of 90, which is like 89.9, which is in quadrant one.

The cosine that is associated with the value of x is positive in quadrant one, so we could say that this will be positive infinity now that The sign of Co is negative in quadrant 2 so it is negative Infinity since these two do not coincide the limit does not exist now let's analyze it graphically the graph of the cosine looks like this. We are going to draw it with dotted lines, that is the secant cosine function which is 1/cosine, you can graph it this way wherever there is where the graph of the currency touches the xais draw a vertical line, it will be a vertical ASM at that point now to graph the secant function simply draw the reciprocal of the cosine function so now we have to look at the values ​​of Pi is in the middle, this is 3 pi over 2, but we need to focus on pi/2, so what is the limit as X? approaches pi/ 2 from the left 89.9 so to speak, as we approach pi/ 2 from the left, notice that the curve if we follow it goes up to positive Infinity, which is the answer we got.

Cinema 89.9 is positive because it is in quadrant one. so the left side was positive now the right side if we follow it when negative, however, these two limits do not coincide, therefore the general limit does not exist. Try these sin 3x over 5x 2x over sin 7x and sin 9x over 8X if you see a question like this, what is the answer if you have multiple choice tests? and you want to know what the answer is the first is simply 3 over 5 the second is 2 over 7 the last is 9 over 8 you can use direct substitution to prove it so make sure your calculator is in radian mode and let's plug in a number which is very close to zero so let's choose 0.1 so S 3 * 0.01 / 5 multiplied by 0.01 if you write this you should get 59 9 91 which is approximately 6 6 if you convert it to a fraction multiply up and down by 10 which is 6 over 10 and if you divide the top and bottom by two you get 3 over 5 so 3 over five is the same as 6 so direct substitution always works if you enter a number very close to zero this is how you can quickly Get the answer, but sometimes in a test they may want you to justify the answer, so let's justify the first one.

Now there is a rule that you need to know and here is the limit when and down by three. Now 5 * 3 is the same as 3 * 5, both expressions are equal to 15. I can change the five and three so that the limit when two are equal, you can see that all of this will become one, so it's 1 * 3 5 but to show your work, we're not done yet, so what we're going to do is replace 3x with Y, so we're going to say that Y is equal to 3x, therefore, when y/y * 3 5, now we can see that this combined expression is equal to 1, so it is 1 * 3 over 5 and the final answer is 3 over 5 now, sometimes you may see a tangent to the problem, so what is the limit as I'll write it as 4X over 1 * 1/ tangent 4X * 4 4, so this expression is the same as now 4X over tangent X tangent is s over cosine, so 1 over tangent is cosine over s, so above I have cosine 4X at the bottom without 4X times, this is supposed to be a five, not a four, so multiplied by 5 over 4, now I'm going to take this and move it here so now we have 4X / sin x time cosine 4X over 1 oh, this is sin 4X by the way, let's not forget that now we are going to say that Y is equal to 4X and as y over sin y, while this 4X is now also y, so cos y * 5 over 4, so when y approaches zero and /sin Y is 1, then this whole expression will be one now the limit when and approaches zero for cosine Y will be cosine 0 and then multiplied by 5 over 4 cosine 0 is 1, so our final answer is 5 over four.

Here is another trigonometric problem: what is the limit as X approaches zero for function 1? - x / cosine ? As X approaches zero for the Ln X function, what is the answer? So let's check the left side and the right side. So what is the limit as X approaches zero from the left side? So what is Ln .01 if you plug Ln ne01 into the calculator it will say error you can never have a negative inside a natural log it doesn't work it is not part of the domain for natural logs so the left side of the limit is not exists, the right side of the boundary does exist if you connect it.

At 1 you will get a negative number, it will be -2.3 if you connect Ln 01, it is 4.6, if you connect Ln 0000001, it is -3.8. Notice that the negative number is increasing, it is becoming more negative and even though it is doing so. at a slow pace you will eventually approach negative Infinity as the left side and right side do not match. The limit does not exist. Now let's analyze this function graphically. The natural logarithmic function has a vertical ASM toote at x equals z and is an increase. function, but it increases at a slower and slower rate, so as you can see, there is nothing on the left side, so the limit as X approaches zero on the left does not exist because there is nothing there except the limit when X approaches zero from the right. exists because as you follow the curve towards zero from the right side, it goes down to negative infinity, so only the right side of the limit exists for the natural log in this particular example, so what is the limit when is close to 5 for the lnx - 5 function? and also find the left side limit and the right side limit if you graph the function, the vertical ASM toote is now at positive 5, you have shifted five units to the right to find a vertical ASM toote, set the inside part equal to zero and solve for x you will get that but the right side as we get closer to five from the right goes. up to Negative Infinity now, since the left side doesn't exist, the overall limit doesn't exist.

If you plug 5.1 into logarithm, so try this: what is the limit as X approaches 4 for Ln 4 - x, so the limit as negative number, so it won't exist, but if we enter four from the left side, let's say 3.9 4 minus 3.9 is a positive number, so it will exist and then this is the one that's going to be negative infinity now if you graph it by the way, this limit does not exist, the vertical ASM toote is at four if you set 4 - xal a0 x is 4, but notice that we have a negative sign in front of the , it won't go this way, but since it's negative, graphically or you can enter numbers so that when we approach four from the left side, it goes down to negative Infinity, what is the limit as X approaches zero? the function e to 2 * without? 1 /x for this type of problem you need to use something called The Squeeze Term, so let's say if you have three fun functions where G is greater than F but less than H and let's say if you know that the limit as X approaches zero for f let's say there are three and the limit as X It should also be equal to three, so that is the main idea behind the compression term, so we need to create two functions, so let's say g of x, the medium is x^2 without x without * 1x now , whenever you graph a sine wave it varies between 1 and negative 1 it will always vary between those two numbers so sometimes s will be a negative 1 and other times s will be a positive one so the lowest value we can get for this function is x2 because the lowest sign it can be is negative 1, the highest sign it can be is positive, so we can say that the lowest value of the function is negative x^2 and the highest value is x s positive, so this function is between x^2 and positive x^2, so let's find the limit as that if you enter zero you will get zero and then the limit which is the top of the function the limit as for the intermediate function that is between x^2 and x^2, this should also be equal to zero and that is the main idea behind compression.

You can usually identify it because you will see s or cosine multiplied by some monomial. Try this for the limit as that's the lowest the cosine could be, so it's negative X Cub, thisanswer for this? Well, if we substitute Infinity into the equation we get e to the power of infinity now let's say that if you plug th into the equation e to the power a00 is a very large number, so it will approach positive Infinity. Now what happens to the limit as X approaches negative Infinity? So let's use substitution. What is negative infinity? This is the same as 1./e to Positive Infinity as long as you have a negative exponent, if you move the E up and down the negative sign will change and become positive.

Now we said that e at infinity is infinity, so we have 1 over infinity 1. / a big number is a small number, so the answer is zero, now we can prove it based on the graph of e DX, the horizontal ASM toote is the x-axis and the function increases at an increasing rate, so as the limit as X approaches Infinity for the function E to X is infinite. Now, what happens to the leftmost behavior as X approaches negative? Infinity, what's going to happen? So if we follow the curve to the left side, notice that it approaches the x-axis where Y is equal to zero, so the behavior on the left end is zero but the behavior on the right end is infinite, what is the limit?

X approaches positive Infinity for the inverse tangent function. What do you think is the answer? Let's find out what is the inverse tangent of a large number equal to the inverse tangent of a th. Let's make sure the calculator is in radian mode, so the inverse tangent actually let's keep it in degree mode the inverse tangent of a th000 is about 9.94 now what about the inverse tangent of let's say a million or 1*10 6? This will be 89999 9°, therefore the inverse tangent as Pi/2 from the left side for the tangent function is positive infinity. The reason I chose the left side is because if you plug in an inverse tangent of th000 we get 89.9 which is less than 90 or to the left of 90, since these two are tangents and inverse tangents to each other we need to change the values X and Y, what happens now? the limit as X approaches Negative Infinity for the inverse tangent function, so what happens if we enter a very large negative number?

So what is the inverse tangent of a000? It will be 89.9, so it is now close to Pi/2 if you graph the inverse. tangent function looks like this there is a horizontal ASM toote at pi/ 2 and negative pi/ 2 the tangent function has a vertical ASM toote active piun 2 and pi/ 2 but the inverse tangent function has a horizontal ASM toote at those two points changes quite a bit, so to speak, now the graph looks like this, so notice that as X approaches the positive infinity, then if you go to the right and notice that the other horizontal toote asm that is Pi/ 2, so you can always find the answer if you know the shape of the graph.

Now let's say if you have a function f ofx is equal to 3x. + 2 and if you are asked to find the points of discontinuity, how would you do it if you have a fraction equal to zero, so this is the vertical ASM toote, so X cannot be equal to -2? You see the graph of this function. so there is a horizontal ASM toote at yal 0 since this function is bottom heavy and is vertical at -2 and looks like this, so if you can't cancel the expression X+ 2 it is a vertical ASM toote and therefore you have an infinite discontinuity which is a non-removable discontinuity now what about this function?

Let's say if f ofx is equal to 2 * equal to z x cannot equal 1 and X cannot equal -2 now how would you classify each discontinuity point? Notice that we can cancel X x - one, so this is a hole or what is a point. discontinuity the hole is also known as removable discontinuity the other two factors that do not cancel are the vertical ASM tootes and they can look like this, they can go in opposite directions or in the same direction but it is an infinite discontinuity so how could we not cancel x y x + 2 It is associated with a vertical asmt, so it is a non-removable discontinuity.

Now what if you have an

absolute

value function? Let's say the

absolute

value of X+ 2 over x + 2. Now notice that we can cancel 2 and x + 2, so it's x + 2 when I take it back, this has to be greater than -2 and for this one it is less than -2, so if you take the limit as X approaches -2 from the left side, we need to use the negative version of the absolute value function, so it will be x + 2 / x + 2 which will be 1 and if you take the limit when 2/by where is the discontinuity point so when you graph the function function it will look like this to the left of -2 it will actually be negative 1 and to the right of -2 it will be either positive 1 or negative -2 that's where it changes direction , but notice that we have a jump discontinuity, so whenever you have an absolute value over that same function it is a jump discontinuity which is also known as a non-removable discontinuity for those of you who are wondering why we can divide the function of absolute value in two parts X positive and X negative.

Here's why consider the graph of the absolute value of x. It's a V shape. Now this part of the graph is equal to X. This part of the graph has a negative value. slope is decreasing this is y is equal to x so as you can see when X is less than Z it is equal to POS a symbol, but I'll let you put it there, just know that when you have an absolute value function you can split it into X or negative at x + 4 and x + 4, so it will change direction at 4, that's where the inside part is equal to zero, so it will be the positive version when negative part of the function.