# What is the limit of (3x^2-8x+1)/(2-7x^2) as x goes to infinity?

Sep 26, 2015

${\lim}_{x \rightarrow \infty} \frac{3 {x}^{2} - 8 x + 1}{2 - 7 {x}^{2}} = - \frac{3}{7}$

#### Explanation:

An attempt to use the quotient rule for limits at infinity gets us the indeterminate form $\frac{\infty}{\infty}$.

However, $\frac{3 {x}^{2} - 8 x + 1}{2 - 7 {x}^{2}}$ can be rewritten for $x \ne 0$.
(when we ask about limits at infinity, we can safely ignore what happens when $x = 0$)

The rewritten fraction will have a denominator with a finite limit.

There are at least three ways to describe this rewriting.

1) Divide the numerator and denominator by the greatest power of $x$ in the denominator. In this case divide both by ${x}^{2}$

2) Multiply numerator and denominator be $1$ over the greatest power of $x$ in the denominator. In this case multiply both by $\frac{1}{x} ^ 2$

3) Factor out the greatest power of $x$ in the denominator from both the numerator and the denominator and then reduce the fraction.

I learned and am still most comfortable with description 3).

$\frac{3 {x}^{2} - 8 x + 1}{2 - 7 {x}^{2}} = \frac{{x}^{2} \left(3 - \frac{8}{x} + \frac{1}{x} ^ 2\right)}{{x}^{2} \left(\frac{2}{x} ^ 2 - 7\right)}$

$= \frac{\cancel{{x}^{2}} \left(3 - \frac{8}{x} + \frac{1}{x} ^ 2\right)}{\cancel{{x}^{2}} \left(\frac{2}{x} ^ 2 - 7\right)}$

${\lim}_{x \rightarrow \infty} \frac{3 {x}^{2} - 8 x + 1}{2 - 7 {x}^{2}} = {\lim}_{x \rightarrow \infty} \frac{3 - \frac{8}{x} + \frac{1}{x} ^ 2}{\frac{2}{x} ^ 2 - 7}$

$= \frac{3 - 0 + 0}{0 - 7} = - \frac{3}{7}$

Remember the key idea is to rewrite so that the denominator has a finite limit.

Example 1
${\lim}_{x \rightarrow \infty} \frac{2 {x}^{5} + 4 {x}^{2} - 9}{3 {x}^{5} + 6 x + 1} = {\lim}_{x \rightarrow \infty} \frac{\cancel{{x}^{5}} \left(2 + \frac{4}{x} ^ 3 - \frac{9}{x} ^ 5\right)}{\cancel{{x}^{5}} \left(3 + \frac{6}{x} ^ 4 + \frac{1}{x} ^ 5\right)} = \frac{2}{3}$

Example 2
${\lim}_{x \rightarrow \infty} \frac{{x}^{3} + 2 {x}^{2} - 7 x + 6}{{x}^{2} + x + 9} = {\lim}_{x \rightarrow \infty} \frac{\cancel{{x}^{2}} \left(x + 2 - \frac{7}{x} + \frac{6}{x} ^ 2\right)}{\cancel{{x}^{2}} \left(1 + \frac{1}{x} + \frac{9}{x} ^ 2\right)}$ $\text{ }$ which has the form $\frac{\infty + 2}{1}$, so we write

${\lim}_{x \rightarrow \infty} \frac{{x}^{3} + 2 {x}^{2} - 7 x + 6}{{x}^{2} + x + 9} = \infty$

Example 3

${\lim}_{x \rightarrow \infty} \frac{4 {x}^{2} + 2 x - 1}{2 {x}^{3} - 7 {x}^{2} + 8} = \text{ what?}$

find the greatest power of $x$ in the denominator.

factor it pout of the numerator and the denominator, and reduce

so the denominator no longer has an infinite limit

the limit of the new denominator is $2$

what is the limit of the new denominator?

If you said $0$, good

Now what is $\frac{0}{2}$?

If you said $0$ again, you're correct. So the limit is $0$
Take a break and have a cookie! Good job.