Find all vertical asymptotes

Question

Find all vertical asymptotes $x=a$ of the following functions. For each value of $a$ , determine ${\displaystyle\lim_{x\to a^{+}}}f\left(x\right)$ , ${\displaystyle\lim_{x\to a^{-}}}f\left(x\right)$ , and ${\displaystyle\lim_{x\to a}}f\left(x\right)$ .
$f (x) = \frac{x + 1}{x^{3} - 4 x^{2} + 4 x}$

$f (x) = \frac{x + 1}{x^{3} - 4 x^{2} + 4 x}$

Accepted Answer

Welcome back, everyone. In this problem, we want to determine all vertical asymptotes X equals A of the function F X equals 2 X + 3 divided by X the 4th minus 8 X cubed plus 16 X2. Evaluate the limit of FX as X approaches A from the right as it approaches A from the left, and as it approaches A for each value of A. Now, let's figure out the first part of our problem. Let's try to find all vertical asymptotes for the function F of X. Remember that a function has a vertical asymptote at a point where the denominator is 0 and the numerator is not 0. So let's set the denominator equal to 0, solve for X, and check if those values make the numerator equal to 0, OK? So setting the denominator to 0. OK, for function FFX. Then we're going to get X to the 4th minus 8 x cubed plus 16 X squad to be equal to 0. Here we can factor out X2 and be left with X2 minus 8 X + 16 equals 0, and we could factor X2 minus 8 X + 16 even further because it's the perfect square. And when we do that, we're going to be left with X2d multiplied by X minus 4 squared being equal to 0. Now what does this tell us? Well, this tells us that either X2 equals 0. Well, that X2 equals 0 and X minus 4 all 2 equals 0, OK. And if X squared is 0, that means X equals 0. And if X minus 42 equals 0, that means X must be equal to 4. If we square root both sides and add 4, OK. So X equals 0 and X equals 4 are the values that we make our denominator 0. Now of those two values, which one of those will make our numerator 0? Let's go to the right here. Now let's start with X being 0. At X being equal to 0, our numerator, 2 x + 3 is going to become 2 multiplied by 0 + 3 and 2 multiplied by 0 is 00 + 3 equals 3. So since our numerator is not equal to 0, then OK, X equals 0 is, OK, a vertical asymptote. So we found one of the vertical asymptotes for FFX. Well, let's try it for our next value of X at X equals 4. When X equals 4, then our numerator is going to become 2 multiplied by 4 + 3. That's 8 + 3, which equals 11, and 11 is not equal to 0. Therefore, this also tells us that X equals 4. Is a vertical asympto as well. Now that we have the values for our vertical asymptotes, then we can evaluate the limit for both of those values, OK? As X approaches A from the right, as it approaches A from the left, and as it approaches A, OK? So let's go ahead and see if we can evaluate those. I'm going to put that in red here. Let me start with our point. Well, before we start, OK, for our function FFX. Equal to 2 x + 3. Divided by X to the 4th minus 8 XQ plus 16 X squared. We could rewrite our denominator by factoring it, OK? So that means we could rewrite this as 2 x + 3. Divided by X2, multiplied by X minus 4 R2. It might make it a little bit easier for us to evaluate our limits. OK. Now, when A equals 0, let's start with that value, OK. Then for the numerator, 2 X + 3, as X approaches 0, OK, from both sides, 2 X + 3. Is going to approach 0. Is it going to approach 3? Sorry, OK, because as X approaches 0, then this will become this term 2 X will become 0, and we're going to be left with 3, OK? And now for the denominator as X approaches 0 from both sides, OK. You know what? I could also write it beside it here. Then X quad will approach 0 from the right, OK, and X minus 4 all 2 will approach 16. Now, the term X2 will always be positive regardless whether X approaches 0 from the right or the left because quaring any real number results in a positive value. OK? Therefore, that means That means the denominator. The denominator. X squared multiplied by X minus 42 is going to approach 0 from the right multiplied by 16, which is going to approach 0 from the right. OK. So what does this tell us? Well, as we said, X2 will be positive regardless whether X approaches 0 from the left or the right. So from this, it can be concluded that the limit, OK. As X approaches 0 from the right of FF X is going to be equal to the limit as X approaches 0 from the left of FF X, and they're both gonna be equal to the limit as X approaches 0 of FF X, which is going to be 3, what the numerator approaches. Divided by 0 from the right what the denominator approaches, and this is going to be equal to infinity. Thus, the solution to all of our limits are infinity and there is a vertical asymptote at X equals 0 and X equals 4. Thanks a lot for watching everyone. I hope this video helped.

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