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).

$f (x) = \frac{4 x^{3} + 1}{2 x^{3} + \sqrt{16 x^{6} + 1}}$

Accepted Answer

Welcome back everyone. In this problem, we want to find the following limits and determine the horizontal asymptotes if any for the function G of X equals six X cubed plus five divided by two XC plus the square root of 16 X to the sixth plus seven. The limit as X approaches infinity of G of X and the limit as X approaches negative infinity of G of X A says the limit as X approaches infinity of G of X is negative three as it approaches negative infinity is negative three and the horizontal Asymptote is Y equals negative three. B says there are one and the negative three respectively and the horizontal asymptotes are Y equals one and Y equals negative three. C says there are one and the negative three respectively and there is no horizontal Asymptote. And D says they are one and one and the horizontal Asymptote is Y equals one. Now, how are we going to find the following limits for our function G of X? Well, making a note that if X equals zero, sorry, if X is positive, that is X is greater than zero. The root of X to the sixth equals X cubed. And if X is less than zero, the root of X to the six equals negative X cubed, we can keep that in mind as we evaluate our limits as X approaches infinity and negative infinity respectively. OK? And then those values will help us to figure out what the horizontal Asymptote is. So let's start with the limit as X approaches infinity. OK. Let me put that in red here. Now for the limit as X approaches infinity of G of X, OK. Then we know that we're looking for the limit as X approaches infinity. OK? Of six XQ plus five divided by two XQ plus the square root of 16 X to the sixth plus seven. But not to make it easier to solve, we can try to simplify our expression. First, let's try to divide through by X cube. OK. Now when we do that, OK, then we're going to have G of X equal to six X cubed divided by X cubed plus five divided by X cubed all divided by two X cubed divided by X cubed plus the square root of 60 X to the sixth. Now remember X is greater than zero here because it's approaching infinity. So we're going to divide by a positive value of X cubed and a positive value of X cube can also be written as the square root of X to the sixth. So under a square root, we can have this as 16 X to the sixth, divided by X to the sixth plus seven divided by X to the sixth. Now, when we go ahead and divide through here for G of X six, X cubed divided by X cubed equals six, we're gonna have plus five divided by X cubed. As the other term in our denominator, it's gonna be two because X cubed cancels X cubed plus the square root of 16 plus seven divided by X to the sixth. And now this is going to be our value of G of X as X approaches infinity. Now that it looks a bit simpler, we can evaluate the limit. OK. Because now this tells us then that the limit as X approaches infinity of G FX K is going to be equal to six plus five divided by X cubed. Now remember that five divided by X cubed is going to approach zero as X approaches infinity. So that's really six plus zero divided by two plus the square root of 16 plus zero because seven divided by X to the sixth will approach zero as X approaches infinity. Now, when we simplify that, that's going to be six divided by two plus the square root of 16, the square root of 16 is 42 plus four equals six and six divided by six equals one. Therefore, the limit as the limit of G FX as X approaches infinity equals one. And since the limit is one that means the horizontal Asymptote, OK. The horizontal Asymptote is also going to be oh sorry, the value is going to be Y equals one. Yeah. As X approaches infinity, no, now that we have that OK, then we can look at the next problem that is we can try to figure it out for the limit as X approaches negative infinity. OK. So next, OK. The limit as X approaches negative infinity of G FX for this limit. Now we know that we can divide every term by X cubed and because X is approaching negative infinity, X will be negative. Thus, the radical expression can be divided by negative X cubed. OK. So what this means then is that we can rewrite OK. G FX as six X cubed divided by X cubed plus five, divided by X cubed all divided by a two X cubed divided by X cubed minus the square root. OK? Of 16 X to the sixth divided by X to the sixth plus seven divided by X to the sixth. And again, this is going to look similar to our previous problem. But no, it's just that it's going to turn out to be six plus five divided by X cubed divided by two minus the square root of 16 plus seven divided by X to the sixth. Now we have an expression for G of X that means our function or sorry. Our limit is going to be the limit. OK. As X approaches negative infinity of six plus five divided by X cubed divided by two minus the square root of 16 plus seven divided by X to the sixth. Again, as X approaches negative infinity five divided by X cubed and seven divided by X to the sixth will approach zero. And thus that will give us six divided by two minus the square root of 16, which is six divided by two minus 42 minus four equals negative two and six divided by negative two equals negative three. So the limit of F FX as X approaches sorry, the limit of G of X as X approaches negative infinity equals negative three. And now, since our limit is equal to negative three, then the function has a horizontal asym to it at Y equals negative three. OK. As X approaches negative infinity. Thus, that tells us that the limit as X approaches infinity of G of X is one. The limit as X approaches negative infinity is negative three and our horizontal asymptotes are Y equal one and Y equals negative three. If we go back to our answer choices that tells us that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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