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

Oct 4, 2016

0

#### Explanation:

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

$= {\lim}_{x \to \infty} \sqrt{x} \sqrt{1 - \frac{2}{x}} - \sqrt{x}$

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

expanding by Binomial Expansion

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

$= {\lim}_{x \to \infty} - \frac{1}{\sqrt{x}} + O \left(\frac{1}{x} ^ \left(\frac{3}{2}\right)\right) = 0$

Oct 5, 2016

$0$

#### Explanation:

${\lim}_{x \rightarrow \infty} \sqrt{x - 2} - \sqrt{x}$

We can rewrite this square root using its conjugate:

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

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

We can try to take a factor from the denominator:

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

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

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

Notice that as $x \rightarrow \infty$, we see that $\frac{2}{x}$ goes to $0$. So, we can "evaluate" this limit:

$= {\lim}_{x \rightarrow \infty} \frac{- 2}{\sqrt{\infty} \left(\sqrt{1 - 0} + 1\right)}$

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

$\sqrt{\infty}$ is still infinite, so just like how $\frac{2}{x}$ approached $0$, this will also tend towards $0$.

$= 0$