# How do you find the limit of sqrt(x^2 + 1) - x as x approaches infinity?

Aug 24, 2016

$= 0$

#### Explanation:

${\lim}_{x \to \infty} \sqrt{{x}^{2} + 1} - x$

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

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

$= {\lim}_{x \to \infty} x \left(\left(1 + \frac{1}{2} \cdot \frac{1}{x} ^ 2 + O \left(\frac{1}{x} ^ 4\right)\right) - 1\right)$

$= {\lim}_{x \to \infty} x \left(\frac{1}{2 {x}^{2}} + O \left(\frac{1}{x} ^ 4\right)\right)$

$= {\lim}_{x \to \infty} \frac{1}{2 x} + O \left(\frac{1}{x} ^ 3\right)$

$= 0$

Aug 24, 2016

Rewrite the expression using the conjugate.

#### Explanation:

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

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

As $x \rightarrow \infty$, the denominator also increases without bound, so the ratio goes to $0$.

The following is optional.

From $= \frac{1}{\sqrt{{x}^{2} + 1} + x}$, we could continue:

For $x > 0$, we get

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

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

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

As $x$ increases without bound, $\left(\sqrt{1 + \frac{1}{x} ^ 2} + 1\right) \rightarrow 2$, and $x \rightarrow \infty$, so

$\frac{1}{x \left(\sqrt{1 + \frac{1}{x} ^ 2} + 1\right)} \rightarrow 0$