# How do you find the limit (sqrtx-1)/(x-1) as x->1?

Oct 29, 2016

The limit is $\frac{1}{2}$

#### Explanation:

We use l'Hôpital's Rule

Limit $\frac{\sqrt{x} - 1}{x - 1} = \frac{0}{0}$
$x \to 1$
So this is impossible, so we apply l'Hôpital's Rule

Limit $\frac{\sqrt{x} - 1}{x - 1} = \frac{\left(\sqrt{x} - 1\right) '}{\left(x - 1\right) '} = \frac{1}{2 \sqrt{x}} \cdot \frac{1}{1} = \frac{1}{2}$
$x \to 1$

Oct 29, 2016

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

#### Explanation:

Note that:

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

Hence:

${\lim}_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} = {\lim}_{x \to 1} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{x} - 1}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(\sqrt{x} - 1\right)}}} \left(\sqrt{x} + 1\right)}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}} = {\lim}_{x \to 1} \frac{1}{\sqrt{x} + 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}} = {\lim}_{x \to 1} \frac{1}{1 + 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}} = \frac{1}{2}$

Oct 29, 2016

$\frac{1}{2}$

#### Explanation:

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

so

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