# How do you find lim (sqrtx-1)/(x-1) as x->1^+ using l'Hospital's Rule or otherwise?

Dec 9, 2016

I wouldn't use l'Hospital for this.

#### Explanation:

I would factor: $x - 1 = \left(\sqrt{x} + 1\right) \left(\sqrt{x} - 1\right)$

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

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

(Of course, I wouldn't write all that if I wasn't trying to explain.

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

OR

"Rationalize the numerator by multiplying by $\frac{\sqrt{x} + 1}{\sqrt{x} + 1}$ to get

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