Determine

Question

Determine $\lim_{x\rightarrow\infty}f\left(x\right)$ and $\lim_{x\rightarrow-\infty}f\left(x\right)$ for the following functions. Then give the horizontal asymptotes of $f$ (if any).

$f (x) = \frac{\sqrt[3]{x^{6} + 8}}{4 x^{2} + \sqrt{3 x^{4} + 1}}$

Accepted Answer

Welcome back everyone. In this problem, we want to calculate the following limits and determine the horizontal asymptotes. If any for the function F FX equals the cube root of X to the sixth plus 27 divided by five X squared plus the square root of two X to the fourth plus three. The limit of FX as X approaches infinity, the limit of FX as X approaches negative infinity. A says the limit as X approaches infinity and the negative infinity and the horizontal asym Y is equal to one divided by five plus the square root of two. B says the limit as X approaches infinity is one divided by five plus the square root of two as it approaches negative infinity is one divided by five minus the square root of two. And there is no horizontal Asymptote. C says they are all 1/5 and D says the limit as X approaches infinity is 1/5 as it approaches negative infinity is negative 1/5. And the horizontal asymptotes are Y equals 1/5 and Y equals negative 1/5. Now, how are we going to figure out the limits and determine the horizontal asymptotes if any, well, one thing that we can note is that for all values of X, the cube root of X to the sixth is equal to X squared and the square root of X to the fourth is equal to X squared. And that's going to come in a bit handy when we start to figure out or simplify our function to find the limits. Now to figure out what our limits are. Let's start with the limit. Let's put that in red here. The limit as yeah, you know what? Let me, let me just indicate here, this is the first thing we're doing. So solving for the limit as X approaches infinity of F of X. Now when we look at F FX, it looks a bit complicated here, but we can simplify, simplify it by dividing through by the smallest power or the smallest degree in our expression, which is going to be the squared term. So we can divide every term by X squared. Now what that means? OK, let me get rid of this here. That means then that F FX is going to be equal to the cube root. Remember we can rewrite X squared as the cube root of X to the sixth. So that's the cube root of X to the sixth, divided by X to the sixth plus 27 divided by X to the sixth. I know this is all being divided by five X squared divided by X squared plus the square root of two X to the fourth divided by X to the fourth plus three, divided by X to the fourth. Now, when we start to simplify, we're going to get FFX to be equal to the cube root of one plus 27 divided by X to the sixth. And that's all divided came by five plus the square root of two plus three divided by X to the fourth. So this is a simpler way for us to write F FX now that we have FX. This way we can go ahead and solve for the limit as X approaches F FX. That is the limit. As sorry as X approaches infinity, my apologies of F FX because now OK. Then that means here as it approaches as X approaches infinity, we know that 27 divided by X to the sixth and three divided by X to the fourth will be, will approach zero rather. So really then this limit is going to be equal to the cube root of one, divided by five plus the square root of two and the cube root of one equals one. OK. That's one divided by five plus the square root of two. No, that's our limit as X approaches infinity. What about our limit? As X approaches negative infinity of F of X? Well, the calculation as X approaches negative infinity is going to be the same. OK. In other words, we know that this is going to be equal to one divided by five plus the square root of two. Now notice here that we have both of our limits. OK? Both of our limits as X approaches infinity and negative infinity equal to one divided by five plus the square root of two. OK. So since the limit of F FX as X approaches infinity is equal to the limit of F FX as X approaches negative infinity. OK. Then that tells us that the given function as a horizontal Asymptote that is equal to that value. In other words, the horizontal Asymptote is going to be Y equals one divided by five plus the square root of two. Therefore, this value is the limit as X approaches infinity as X approaches negative infinity. And it is the value of the horizontal Asymptote. Y. Therefore, that tells us then when we look back at our answer, choices that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Determine $\lim_{x\rightarrow\infty}f\left(x\right)$ and $\lim_{x\rightarrow-\infty}f\left(x\right)$ for the following functions. Then give the horizontal asymptotes of $f$ (if any).

$f (x) = \frac{\sqrt[3]{x^{6} + 8}}{4 x^{2} + \sqrt{3 x^{4} + 1}}$

Key Concepts

Limits at Infinity

Horizontal Asymptotes

Rational Functions

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