# How do you find lim sqrt(x+2)-sqrtx as x->oo?

Jan 22, 2017

${\lim}_{x \rightarrow \infty} \left(\sqrt{x + 2} - \sqrt{x}\right)$ has indeterminate initial form $\infty - \infty$. See below.

#### Explanation:

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

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

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

As $x \rightarrow \infty$, the denominator $\sqrt{x + 2} + \sqrt{x} \rightarrow \infty$

Therefore,

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

$= 0$