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).

$f (x) = \frac{6 x^{2} - 9 x + 8}{3 x^{2} + 2}$

Accepted Answer

Welcome back, everyone find the following limits and identify the horizontal asymptotes. If any for the function G of X is equal to four X squared plus five X minus three, divided by two X squared minus seven, we want to evaluate two limits limit of G of X one X approaches positive infinity and negative infinity. There are also four answer choices. ABC and D, they give us the answers to the two limits and possible horizontal asymptotes. So essentially, we want to identify the two limits. Let's begin with the first one limit when X approaches infinity of G of XG of X is given. So we are going to substitute and we're going to write it as four X squared plus five X minus three. We are dividing that by two X squared minus seven. If we directly substitute infinity, we're going to get infinity divided by infinity, right? Essentially this is an indeterminate form. And we are going to actually identified the limit by dividing both sides of our rational function by X squared because this is the variable with the highest exponent. So essentially we end up with limit when X approaches infinity. Now four X squared divided by X squared, gives us 45 X divided by X squared, gives us five divided by X. And then we are subtracting three divided by X squared. Now, in the denominator, we have two X squared which is going to be divided by X squared. This gives us two and then we are subtracting seven divided by X squared. Now, if we look at this expression, we have 45 divided by infinity is going to give us zero and three divided by infinity squared, which is three divided by infinity is zero. So we're subtracting zero. In the denominator, we have two minus seven divided by X squared. So that's seven divided by infinity which is zero four divided by two, gives us a limit of two. Now does anything change when we evaluate the limit when X approaches negative infinity? Well, essentially we're just subtracting zero, then we are subtracting zero once again. And in the denominator, we are subtracting zero, nothing really changes. The reason why nothing changes is simply because those rational functions where we have a number divided by infinity still approach zero. It doesn't really matter from the left or from the right because we have whole numbers four divided by two and the ratio will still be two. So whether we approach infinity as positive infinity or negative infinity, we will get the same limit of two. Meaning there is only one horizontal Asymptote. Let's recall that horizontal asymptotes they have an expression of Y equals a number. And specifically, that number is equal to the limit itself whenever we approach infinity or negative infinity. And since for both limits, we have two, there's only one horizontal Asymptote in a form of Y equals two. If we look at the answer choices, the correct answer to this problem would be option B limits are equal to two and the horizontal asym code has a form of Y equals two. Thank you for watching.

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