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).

$f (x) = \frac{3 x^{3} - 7}{x^{4} + 5 x^{2}}$

Accepted Answer

Welcome back, everyone, calculate the following limits and identify the horizontal asymptotes. If any for the function H of X is equal to two X cubed minus nine divided by X four plus or X squared, we want to evaluate two limits limit of H of X one X approaches positive and negative infinity. This is the definition of a possible horizontal Asymptote, right. So essentially we're given four answer choices. ABC and D those give us the answers to the, to the two limits and the expressions of the horizontal asymptotes. What we want to do is basically evaluate the two limits. Let's begin with the limit. When X approaches positive infinity of H of XH of X is to X cubed minus nine divided by X to the power of fourth plus four X squared. If we directly substitute X equals infinity, while we're going to end up with infinity divided by infinity. And this is an indeterminate form. So if we have a irrational function, one of the ways to evaluate the limit is to divide both sides by the highest exponent of X. Essentially, if we take the variable X, we can identify that the highest exponent of X is the fourth exponent right, or the fourth power. And we want to divide all of the terms by X to the power of four because this is the variable with the highest exponent. So we end up with limit when X approaches infinity, we want to divide two X cubed by X, the power of fourth. And this gives us two divided by X and the numerator, we are subtracting nine which is going to be divided by X to the power of fourth. And then our denominator has X to the power of four divided by X to the power of fourth, which is one. And when we divide four X squared by X fourth, we get four divided by X squared. Now if we substitute X equals infinity or basically X approaches infinity, that would be more accurate. We get two divided by infinity which approaches zero. We are subtracting nine divided by infinity, right, which is zero once again. And then in the denominator, we have one plus four divided by infinity. This approach is zero. So zero divided by one gives us zero. What if X approaches negative infinity, nothing really changes, right? We simply get those zeros such as two divided by X, right. And then in this case, we would approach zero from the left hand side, but that would still be zero. So we get zero minus zero divided by one plus zero. So this would still give us the same limit. Meaning it doesn't matter if we approach infinity from negative infinity or positive infinity, we still get the same answer of zero. Now, the horizontal Asymptote has a form of Y equals A where A is the number for each limit, the two limits are equal to each other. So we only have one horizontal Asymptote and a form of Y equals zero. If we look at the answer choices, we can see that the correct answer to this problem is option A, the two limits are equal to zero and the horizontal Asymptote is Y equals zero. Thank you for watching.

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