Find all vertical asymptotes

Question

x = a

of the following functions. For each value of

a

, determine

\lim_{x \to a^{+}} f (x)

,

\lim_{x \to a^{-}} f (x)

, and

\lim_{x \to a} f (x)

.

$f (x) = \frac{\cos (x)}{x^{2} + 2 x}$

Accepted Answer

Welcome back, everyone. In this problem, we want to determine all vertical asymptotes X equals A of the function G of X equals sin x divided by X2 minus 7 X. Evaluate the limit of geoX as X approaches A from the right, from the left, and as X approaches A for each value of A. Now, before we can find our limits, let's talk with the first part of our problem. Let's determine all vertical asymptotes for G of X. Now, what do we know about vertical asymptotes? Well, I recall, OK, that if we have a function F of X, that's expressed as a quotient, let's say the quotient is KFX divided by PF X, OK, where KF X is the numerator and the PF X is the denominator. Then a vertical asymptote exists, OK. A vertical asymptote exists. As X approaches some value A. And as FFX approaches infinity or negative infinity, OK. If Or denominator equals 0, OK. And Our numerator is not equal to 0. In other words, if we can for our quotient find values for the denominator being zero where the numerator is not zero, that means a vertical asymptote exists at that point. If not, that is if the numerator is equal to 0, then a hole would be there instead. So let's take our denominator, X2 minus 7 X. Let's find the values which make it zero and then test if those values also make the numerator equal to 0, OK? Now for our denominator, let me put that in 0 here, OK. Uh, setting the denominator. Of GX to 0. OK. Then that means X2 minus 7 X will be equal to 0. If we factor X, that means X multiplied by X minus 7 will be equal to 0, which means then that X equals 0 and X equals 7 are potential vertical asymptotes. Now, let's test the both of them. OK, let me come over to the right here. Now first, let's test when X equals 0. Now when X equals 0, OK, then if we take our numerator. The sin of X, we're gonna get the sin of 0, which equals 0. Since the numerator equals 0, there is no vertical asymptote at X equals 0, but a hole instead, OK? So, there is no vertical asymptote. Now how about when X equals 7? Knowing X equals 7, then we're gonna have the numerator as the sin of 7, which is approximately equal to 0.657. This approximation indicates that the sin of 7 is going to be greater than 0. It's non-zero. Therefore, OK, X equals 7 is a vertical asymptote. So we've solved the first part of our problem. We've phoned or we've determined all the vertical asymptotes for G of X. Next, let's try to figure out the limits of G of X as X approaches A from the left, from the right, and as X approaches A. Where here we know that our value of A is 7. Now, when it, let me put that back in black here. When X, which is a, is equal to 7, OK. Then for all limits, for all limits. The sign of X. For the numerator, the sign of X is going to be approaching the sign of 7, OK? As X approaches 7. And for the denominator, X squad minus 7 X can be factored as X multiplied by X minus 7, OK? And as X approaches 7 from the right, for example, then X multiplied by X minus 7 will also approach 0 from the right, OK? Since X is greater than 7. So starting with our limit from the right, this means, OK, that the limit. Of G of X as X approaches 7 from the right. Is going to be equal to the sign of 7, OK, divided by 0 being approached from the right means getting to a smaller value and the smaller the value that will divide the sign of 7 by then the larger or result, which means that this is going to be equal to infinity. OK, what about from the left? Well, as X approaches 7 from the left, OK. Then X multiplied by X minus 7 will approach 0 from the left, OK? Since X is less than 7. And now in that case, that would mean that the limit, OK, as X approaches 7 from the left of G of X is going to be equal to the sign of 7. Divided by 0 being approached from the left again becoming a very small negative value which is going to make our limit negative infinity. Now that we've found it from the left and the right, what is the limit of G of X as X approaches A, that is 7. Well, The limit as x approaches 74 G of X does not exist. Does not exist. And why doesn't it exist? Well, based on what we know about limits, it doesn't exist because the limit as X approaches several from the left of GFX is not equal to the limit as X approaches 7 from the right of GF X. OK, so because. The left hand limit is not equal to the right hand limit. That means the limit for our value does not exist. Thanks for watching everyone. I hope this video helped.

Watch next