Now let's talk about infinite limits. What is the limit as x approaches zero of the function 1 /x? What's the answer to that? Well, first what we need to do is check the limit at the left side and the right side. So let's start at the left side. What is 1 /.1? 1 /.1 is -10. But now let's plug in a number that's even close to zero from the left side. So it's going to be negative again. Let's try.01. 1 /.01 is00. And if we divide it by.001 that becomes 1,000. Notice that the number gets larger and larger but with a negative sign. So therefore we could say that this approaches negative infinity. Now what about the right side of limit? What is the limit as x approaches zero from the right? 1 over positive.1 is 10. 1 over 01 is 100 and 1 / 01 is positive 1. As you could see the number gets larger and larger. So this is going to approach positive infinity. Now generally speaking whenever you have a small number basically zero from the right side whenever you have a small number in the denominator it makes the value of the fraction large. So this is equivalent to positive infinity. Now keep in mind if it's exactly zero it's undefined. So it really doesn't exist. But if it's close to zero that means it's going to be very large. So whenever you have one and zero from the left, it's going to be negative infinity. So just some things to keep in mind. So now let's finish this question. Since these two limits do not match the left side and the right side. Therefore, the limit as x approaches zero from either side does not exist. Now let's try another problem. What is the limit as x approaches zero from either side of 1 / x^2? So let's evaluate it as x approaches zero from the left side. So if we plug in zero from the left, we're still going to get a small number, but it's going to be positive. So you can treat it as zero from the right and this is going to become positive infinity. So for example, let's say if we plug in.01.01 1^ 2 is actually positive.001 and 1 / that number is positive 10,000. You could type it in your calculator just to confirm it. So that's why we got positive infinity. Now what about the limit as x approaches zero from the right? So if we plug in zero from the right once we square it it's still going to be a positive number. So it's still going to be small. It's going to be very close to zero but it's going to be positive. And this is going to be positive infinity. Therefore the limit I mean excuse me the limit as x approaches zero from the right from either side rather is going to be positive infinity. The number is going to get larger and larger and larger. So that's the answer as x approaches zero from either side. Now let's move on to another example. What is the limit as x approaches 2 from the left of 1 /x - 2? Go ahead and try that problem. Now let's use direct substitution. Let's plug in a number that's less than two. less than two would be like 1.9 1.9 minus 2 is.1 and this is going to be -10. Now if we try 1.99 on the bottom we're going to get.01 and that's going to be 100. So we can see that we're going to get a negative0 or zero from the left on the bottom. So this is going to approach negative infinity and this is the answer for this uh one-sided limit. What is the limit as x approaches 3 from the right of five / x - 3? Let's go ahead and evaluate this one. So three from the right we can plug in 3.1 3.1 minus 3 is 0.1 5 /.1 is 50. So notice that we have a very large positive value. So therefore the limit as x approaches 3 from the right of 5 / x - 3 this is going to be positive infinity. Anytime you get a zero on the bottom, it's either going to be plus or minus infinity. In this case, since we have a zero from the right, it's positive infinity. Now, what about the limit as x approaches -4 from the left of 15 / x + 4? Try that. So, let's plug in a number that's close to4 from the left. So here's4 to the left of it is5. So we want to plug in a number like4.14.1 + 4 that's.1 150 /.1 is 150. So we get a very large negative value. So we could say that this limit approaches infinity. What about the limit as x approaches 2 from the right? Let's say this is x - 2 with a8 on top. What's the answer? So8 / two from the right minus two. This is going to give us zero from the right. So having zero from the right, that's going to be like negative or positive infinity. So it's like8 * positive infinity which in the end will be negative infinity. But let's actually plug in a number to confirm it. So two to the right let's plug in a number that's bigger than 2. So 2.1 could be a good example. 2.1 minus 2 is positive.18 / positive.1 is80. So we have a very large negative number which means the limit approaches negative infinity. Now what about this one? What is the limit as x approaches 1 from either side of 1 / x - 1? So let's evaluate it from the left side first. So as x approaches one from the left, what is it going to be? Well, we can approximate it by plugging a number that's less than one. Let's try 0.9. 9 - 1 that's.1. So this is -10. So we know that this is going to be equal to a large negative number. So we can say that the limit as x approaches one from the left is going to be negative infinity. Now let's check the right side. As x approaches one from the right. So we can approximate using 1.1. So this is going to be positive.1 1 / positive.1 is pos 10. So therefore this limit is going to approach positive infinity. Now these two limits do not match. So this limit does not exist. Here's another one that you can try. The limit as x approaches 2 from the right of x cub / x - 2 raised to the second power. Try that. So we can plug in two on top. On the bottom we need to plug in 2.1. We have to plug in a number that's close to two but to the right of two. 2.1 minus 2. That's going to be like 0.1. 0.1^ squar that's 01. So what we need to take from this is we're going to get a very large positive number. So therefore this is going to approach positive infinity. Let's say if you plug in 2.01 instead 2.01 to the third power that's about 8.12. And if you divide it by 2.01 0 1 - 2^ 2. This is going to be a very large positive number. So in the end we can see that this is going to approach positive infinity. Now what about the limit as x approaches three from either side of this function? Well, if we plug in three, this is we're going to have a zero on the bottom. So, we know it's either going to be positive infinity or negative infinity. So, we just got to check the sign. So, let's start with the right side. On the right side, we need to plug in a number that's bigger than 3. 3.1us 3 is a positive number. and.1^ squar is 01. 1 / 01 is 100 * 8 that's going to be positive 800. So on the right side it approaches positive infinity. on the left side. If we plug in a number that's let's say less than three like 2.9 2.9 minus one I mean 2.9 - 3 is.1 but when you square.1 you're going to get positive.01 so you're going to get the same value positive 800. So therefore these two limits they match. So the limit as x approaches three from either side approaches positive infinity due to the square everything is going to be positive. What is the limit as x approaches -3 from the left given this expression x + 3 / x^2 + x - 6? In this example, the first thing we need to do is factor. What two numbers multiply to -6 but add to the middle coefficient of 1? This is going to be pos3 and -2. So x^2 + x - 6 is equivalent to x + 3 and x - 2. Now if we plug in -3, now -3 + 3 will give us a zero on the bottom. But because it cancels, we can use direct substitution in this problem. So all we got to do is replace x with -3. So it's going to be 3 - 2 which is5. So the answer is 1 /5. Here's another one. What is the limit as x approaches 5 from either side of x^2 / x^2 + 25? Go ahead and work on this problem. Now if we plug in five, we won't get a zero on the bottom. So in this example, we can use direct substitution. 5^ squar or 5 * 5 is 25. 25 + 25 is 50 and 25 / 50 is 12. So this is the answer. Now what is the limit as x approaches zero from the left of the expression 2 - 1 / x? So go ahead and try that. So let's start with this term. What's the limit as x approaches zero from the left of pos 1x? We know that zero from the left this is going to be negative infinity. So if we have negative 1/x this will change to positive infinity. So this becomes 2 + positive infinity which in the end will be positive infinity. And let's try plugging in a small number. Let's try plugging in.01 /.01 that's going to be00. So 2us100 that's going to be 102. So we have a very large positive number which tells us that this limit is going to approach positive infinity. Now what about the limit as x approaches zero from the right of the natural log function ln x? What is the answer? It helps to know the graph of ln x. ln one is zero by the way. It has a vertical asmtote at zero. So it looks like this. So as you approach zero, which is the y ais from the right, notice that it goes down to negative infinity. And if you have your calculator, you can confirm it. Let's say if you plug in ln.1 this is equal to -2.3. Now let's plug in ln.00001. So it has to be from the right side. So it has to be positive. This is going to equal 9.2. So notice that we get a larger negative. Let's plug in a very very small number. 1 * 10^50. So that's like 0.0000 almost 50 times and then a one. This is -15. So slowly becomes more negative as we can see. So that's why it approaches negative infinity. Now what about the limit as x approaches zero from the left of ln x? Notice that there's nothing on the left side. So therefore this limit does not exist. If you were to type in ln.1 in your calculator, you're going to get a math error. That means that it doesn't exist. You can't have a negative number inside a natural log. So therefore the limit as x approaches zero from either side of ln x also does not exist because the left side does not exist but the right side it approaches negative infinity. Now what is the limit as x approaches zero of tangent x? Let's go over some trigonometric functions. Now for this we can use direct substitution tangent of zero is zero. Now let's say if we change it what if x approach p / 2 and we still have the tangent function. What's the answer? Now tangent pi / 2 is undefined. So we can't just use direct substitution. So therefore we need to check the left and right side. So let's start with the right side of tangent. So let's approach two from the right. Pi / 2 is basically equivalent to 90°. So if you want to plug in a number that's close to 90 from the right side, we need to try 90.1. Now feel free to put your calculator in degree mode. tangent 90.1 is about negative 573. If you round it to nearest whole number tan 90.01 is -5,730. So therefore we can see that it's approaching negative infinity. It's becoming more negative as we get closer to 90. And it makes sense because 90.1 is located in quadrant 2 and tangent is negative in quadrant 2. The limit as x approaches pi / 2 from the left. Now let's say if we plug in 89.9 tangent 89.9 is positive 573. So we can see that this is going to approach positive infinity. So these two limits they they don't match. They're different. So therefore the limit as x approaches p /2 of tangent x does not exist. So anytime you see tangent just keep in mind that you can replace it with s / cosine and notice that if you plug in pi / 2 sin p /2 is equal to 1 and cosine pi / 2 is zero. Anytime you get a zero on the bottom it's undefined. So that's a clue that you need to check the left side and the right side of limits as well. So that limit does not exist. Now what about this one? What is the limit as x approaches zero from the right of coseant x? Now keep in mind cosecant is 1 / s and s of 0 is zero. One over 0 is undefined. But since we're approaching zero, chances are our answer is going to be positive or negative infinity. So let's find out which one. Now we plug in a number that's greater than zero since it's to the right of zero. Sine of 0.1 is going to be approximately 0.1 if this is in radian mode. Sine of 0.1 is about 0998 which let's round that to 0.1 and one over.1 is 10. Now if we plug in S of 01 this answer is going to be approximately about 100. So therefore the numbers are getting bigger and bigger as we approach zero. So we can make the conclusion that the limit as x approaches zero from the right of cosecant is positive infinity. Now what about the limit as x approaches pi / 2 of x? Let's try that. So is 1 / cosine and if we plug in pi / 2 cosine p / 2 is zero. So it's undefined. So as we approach pi /2, if we analyze the left and the right side of limits, it's either going to be positive infinity or negative infinity. So let's start with the right side. So cosine of 90.1 is that a positive value or a negative value? 90.1 is in quadrant 2 and cosine is negative. So therefore we can make the conclusion that this is going to be negative infinity. If you plug in one, make sure your calculator is in radian mode or rather degree mode. If you plug in 1 / cosine 90.1, you're going to get -573. Now on the left side of pi / 2, that's 89.9. The angle is in quadrant 1. Cosine is positive in quadrant 1. So we could say this is going to approach positive infinity. So since these do not match, we could say that the limit as x approaches p /2 from either side does not exist. Now here's the last question related to this section. What is the limit as x approaches pos4 from the right of 1 / the x - 2? So if we plug in 4, the of 4 is 2. 2 - 2 is 0. 1 / 0 is undefined. But since it approaches 4 and it doesn't equal 4, it's either going to be positive or negative infinity. So let's substitute x with 4.1. The square root of 4.1 is going to be a number that's just above two. So let's make a number. Let's say 2.1. Let's say if we don't have access to a calculator, it's probably going to be less than 2.1. 1 / 0.1 is 10. So this is a relatively large positive number. The actual value of the square root of 4.1 is 2 02. five and this is actually 40. So this is definitely a bad estimate. But this will lead us to the same final answer and that is since we're getting a large positive number, we could say that the limit approaches positive infinity and that is the answer to the question. So a lot of these problems you can use direct substitution to get the answer. Just make sure you plug in a number that's close to the value that x is approaching and it's always going to work if you do it that way. Now let's talk about how to find the vertical asmtote of a function. So let's say if we have the function 1x + 2. What is the vertical asmtote? To find it set the denominator equal to zero. So the vertical asmtote is going to be x is equal to -2. So let's try some more examples. What is the vertical asmtote in this case? So if we set x - 3 equal to zero, the vertical asmtote is x = 3. Here's another example. Let's say if the function is x - 5 / x - 5 over x + 3. And if we have another one, let's say 7 x + 1 over x - 2 x + 1. What are the vertical asmtotes? So looking at the factor in the bottom x + 1. We just need to change the sign. So the vertical asmtote is going to be x=1. And for the factor x - 2, the vertical asmtote is going to be x= + 2. Simply reverse the sign. Now for the next one for the factor x + 3 it's going to be x= to -3 x - 5 will not lead to a vertical asmtote because it can be cancelled. So x= 5 is a whole everything else are vertical asmtotes. What about this one? 1 / x^2us 4. What are the vertical asmtotes found in this rational function? So what we need to do is factor using the difference of perfect squares technique. The square root of x^2 is x and the roo of 4 is 2. And so it's going to be x + 2 and x - 2. So we can see that the vertical asmtotes are x= -2 and x= pos2. So sometimes you need to factor in order to find it. Here's another example in which you need to factor. So how can we factor x^2 - x - 2? What two numbers multiply to -2 but add up to 1? This is going to be -2 and one. So to factor it, it's simply x - 2 and x + 1. So therefore the vertical asmtotes are x is equal to pos2 and -1. Now for the sake of practice, let's try one more example. Try this one. So take a minute, pause the video and work on it. So we need to factor everything completely. In the numerator we could take out an x which will leave behind xus 2. In the denominator we can factor using the difference of perfect squares. The square root of x 4th is x^2 and the of 16 is 4. Now we can't factor x^2 + 4 without using imaginary numbers. But we can factor x^2 - 4 because we did it in the last example and we know it's x + 2 * x - 2. So this factor leads to a hole. So we have a hole at x = pos2. This one leads to the vertical asmtote which is x= -2. Now let's talk about the other factor that was in the denominator and that is x^2 + 4. Notice that there's no vertical asmmptotes or holes associated with this. If we set it equal to zero, we need to subtract both sides by four. And so x^2 is going to equal4. And if we take the square root of both sides, we're not going to get a real number. Keep in mind the square root of 1 is i. So this is going to be plus or minus the square root of 4 is 2 * the of 1 which is i. So you get posit 2 i and -2 i. These are not real numbers. They're imaginary numbers. So they don't represent any holes or vertical asmtotes. So that's why we could ignore x^2 + 4.