# How do you find the limit of (8x-14)/(sqrt(13x+49x^2)) as x approaches oo?

Jun 19, 2016

Do a little factoring and canceling to get ${\lim}_{x \to \infty} \frac{8 x - 14}{\sqrt{13 x + 49 {x}^{2}}} = \frac{8}{7}$.

#### Explanation:

At limits of infinity, the general strategy is to take advantage of the fact that ${\lim}_{x \to \infty} \frac{1}{x} = 0$. Normally that means factoring out an $x$, which is what we'll be doing here.

Begin by factoring an $x$ out of the numerator and an ${x}^{2}$ out of the denominator:
$\frac{x \left(8 - \frac{14}{x}\right)}{\sqrt{{x}^{2} \left(\frac{13}{x} + 49\right)}}$
$= \frac{x \left(8 - \frac{14}{x}\right)}{\sqrt{{x}^{2}} \sqrt{\frac{13}{x} + 49}}$

The issue is now with $\sqrt{{x}^{2}}$. It is equivalent to $\left\mid x \right\mid$, which is a piecewise function:
$\left\mid x \right\mid = \left\{\begin{matrix}x & \text{for" & x > 0 \\ -x & "for} & x < 0\end{matrix}\right.$

Since this is a limit at positive infinity ($x > 0$), we will replace $\sqrt{{x}^{2}}$ with $x$:
$= \frac{x \left(8 - \frac{14}{x}\right)}{x \sqrt{\frac{13}{x} + 49}}$

Now we can cancel the $x$s:
$= \frac{8 - \frac{14}{x}}{\sqrt{\frac{13}{x} + 49}}$

And finally see what happens as $x$ goes to $\infty$:
$= \frac{8 - \frac{14}{\infty}}{\sqrt{\frac{13}{\infty} + 49}}$

Because ${\lim}_{x \to \infty} \frac{1}{x} = 0$, this is equal to:
$\frac{8 - 0}{\sqrt{0 + 49}}$
$= \frac{8}{\sqrt{49}}$
$= \frac{8}{7}$