# How do you evaluate the limit sqrt(x^2+1)-2x as x approaches oo?

Dec 13, 2016

See below.

#### Explanation:

Change the way the expression is written.

If we try to find the limit as written, we get the indeterminate form $\infty - \infty$.

We could change this to a fraction using $\frac{\sqrt{{x}^{2} + 1} + 2 x}{\sqrt{{x}^{2} + 1} + 2 x}$, but perhaps the following is simpler.

For x !`= 0, we have

sqrtx^2+1) - 2x = sqrt(x^2(1+1/x^2))-2x

$= \sqrt{{x}^{2}} \sqrt{1 + \frac{1}{x} ^ 2} - 2 x$

For positive $x$, $\sqrt{{x}^{2}} = x$ so

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

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

Which has the form $\infty \cdot \left(1 - 2\right)$

Therefore,

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