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### Section 2.7 : Limits at Infinity, Part I

8. For $$\displaystyle f\left( x \right) = \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}}$$ answer each of the following questions.

1. Evaluate $$\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)$$.
2. (b) Evaluate $$\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)$$.
3. Write down the equation(s) of any horizontal asymptotes for the function.

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a Evaluate $$\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)$$. Show Solution

To do this all we need to do is factor out the largest power of $$x$$ that is in the denominator from both the denominator and the numerator. Then all we need to do is use basic limit properties along with Fact 1 from this section to evaluate the limit.

In this case the largest power of $$x$$ in the denominator is just $$x$$ and so we will need to factor an $$x$$ out of both the denominator and the numerator. Recall as well that this means we’ll need to factor an $${x^2}$$ out of the root in the numerator so that we’ll have an $$x$$ in the numerator when we are done.

So, let’s do the first couple of steps in this process to get us started.

$\mathop {\lim }\limits_{x \to \, - \infty } \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\sqrt {{x^2}\left( {\frac{7}{{{x^2}}} + 9} \right)} }}{{x\left( {\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\sqrt {{x^2}} \sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\left| x \right|\sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}}$

Recall from the discussion in this section that.

$\sqrt {{x^2}} = \left| x \right|$

and we do need to be careful with that.

Now, because we are looking at the limit $$x \to - \infty$$ it is safe to assume that $$x < 0$$. Therefore, from the definition of the absolute value we get.

$\left| x \right| = - x$

and the limit is then.

$\mathop {\lim }\limits_{x \to \, - \infty } \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{ - x\sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{ - \sqrt {\frac{7}{{{x^2}}} + 9} }}{{\frac{1}{x} - 2}} = \frac{{ - \sqrt 9 }}{{ - 2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{2}}}$

b Evaluate $$\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)$$. Show Solution

For this part all of the mathematical manipulations we did in the first part up to dealing with the absolute value did not depend upon the limit itself and so don’t really need to be redone here. So, up to that part we have.

$\mathop {\lim }\limits_{x \to \,\infty } \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{\left| x \right|\sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}}$

In this part we are looking at the limit $$x \to \infty$$ and so it will be safe to assume in this part that $$x > 0$$. Therefore, from the definition of the absolute value we get.

$\left| x \right| = x$

and the limit is then.

$\mathop {\lim }\limits_{x \to \,\infty } \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{x\sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{\sqrt {\frac{7}{{{x^2}}} + 9} }}{{\frac{1}{x} - 2}} = \frac{{\sqrt 9 }}{{ - 2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{3}{2}}}$

c Write down the equation(s) of any horizontal asymptotes for the function. Show Solution

We know that there will be a horizontal asymptote for $$x \to - \infty$$ if $$\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)$$ exists and is a finite number. Likewise, we’ll have a horizontal asymptote for $$x \to \infty$$ if $$\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)$$ exists and is a finite number.

Therefore, from the first two parts, we can see that we will get the horizontal asymptote.

$y = \frac{3}{2}$

for $$x \to - \infty$$ and we have the horizontal asymptote.

$y = - \frac{3}{2}$

for $$x \to \infty$$.