# How do you determine the limit of (x)/sqrt(x^2-x) as x approaches infinity?

Apr 26, 2016

${\lim}_{x \rightarrow \infty} \frac{x}{\sqrt{{x}^{2} - x}} = 1$ Bonus: ${\lim}_{x \rightarrow - \infty} \frac{x}{\sqrt{{x}^{2} - x}} = - 1$

#### Explanation:

Note that, for all $x$ other than $0$ (which we are not interested for limits at infinity), we have

$\sqrt{{x}^{2} - x} = \sqrt{{x}^{2} \left(1 - \frac{1}{x}\right)} = \sqrt{{x}^{2}} \sqrt{1 - \frac{1}{x}}$

Note also that $\sqrt{{x}^{2}} = \left\mid x \right\mid = \left\{\begin{matrix}x & \text{if" & x >= 0 \\ -x & "if} & x < 0\end{matrix}\right.$

So we get

lim_(xrarroo)x/sqrt(x^2-x) = lim_(xrarroo)x/(sqrt(x^2)sqrt(1-1/x)

 = lim_(xrarroo)x/(xsqrt(1-1/x) $\text{ }$ (As $x \rightarrow \infty$ we are only interested in positive values of $x$.)

$= {\lim}_{x \rightarrow \infty} \frac{1}{\sqrt{1 - \frac{1}{x}}} = \frac{1}{\sqrt{1 - 0}} = 1$

Bonus

lim_(xrarr-oo)x/sqrt(x^2-x) = lim_(xrarr-oo)x/(sqrt(x^2)sqrt(1-1/x)

 = lim_(xrarr-oo)x/(-xsqrt(1-1/x) $\text{ }$ (As $x \rightarrow - \infty$ we are only interested in negative values of $x$.)

$= {\lim}_{x \rightarrow \infty} \frac{1}{- \sqrt{1 - \frac{1}{x}}} = \frac{1}{- \sqrt{1 - 0}} = - 1$