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

Sep 16, 2015

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

#### Explanation:

I'm not sure the question is typed correctly, so I'll give solutions to both possibilities.

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

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

And

${\lim}_{x \rightarrow \infty} \frac{1}{\sqrt{{x}^{2} + 1} - x} = {\lim}_{x \rightarrow \infty} \frac{1}{\left(\sqrt{{x}^{2} + 1} - x\right)} \frac{\left(\sqrt{{x}^{2} + 1} + x\right)}{\left(\sqrt{{x}^{2} + 1} + x\right)}$

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

$= {\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{2} + 1} + x\right)$

$= {\lim}_{x \rightarrow \infty} \sqrt{{x}^{2} + 1} + {\lim}_{x \rightarrow \infty} x$

$= \infty + \infty = \infty$