# How do you simplify (sqrt(4+h)-2)/h ?

Feb 25, 2017

$\frac{\sqrt{4 + h} - 2}{h} = \frac{1}{\sqrt{4 + h} + 2}$

with exclusion $h \ne 0$

#### Explanation:

Given:

$\frac{\sqrt{4 + h} - 2}{h}$

We find:

$\frac{\sqrt{4 + h} - 2}{h} = \frac{\left(\sqrt{4 + h} - 2\right) \left(\sqrt{4 + h} + 2\right)}{h \left(\sqrt{4 + h} + 2\right)}$

$\textcolor{w h i t e}{\frac{\sqrt{4 + h} - 2}{h}} = \frac{\left(4 + h\right) - 4}{h \left(\sqrt{4 + h} + 2\right)}$

$\textcolor{w h i t e}{\frac{\sqrt{4 + h} - 2}{h}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{h}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{h}}} \left(\sqrt{4 + h} + 2\right)}$

$\textcolor{w h i t e}{\frac{\sqrt{4 + h} - 2}{h}} = \frac{1}{\sqrt{4 + h} + 2}$

$\textcolor{w h i t e}{}$
Footnote

This is the sort of expression you find involved in a limit problem, like:

What is: ${\lim}_{h \to 0} \frac{\sqrt{4 + h} - 2}{h}$ ?

The tricky thing here is that both the numerator and denominator become $0$ when $h = 0$, but we can find the limit using our simplification...

${\lim}_{h \to 0} \frac{\sqrt{4 + h} - 2}{h} = {\lim}_{h \to 0} \frac{1}{\sqrt{4 + h} + 2}$

$\textcolor{w h i t e}{{\lim}_{h \to 0} \frac{\sqrt{4 + h} - 2}{h}} = \frac{1}{\sqrt{4 + 0} + 2}$

$\textcolor{w h i t e}{{\lim}_{h \to 0} \frac{\sqrt{4 + h} - 2}{h}} = \frac{1}{2 + 2}$

$\textcolor{w h i t e}{{\lim}_{h \to 0} \frac{\sqrt{4 + h} - 2}{h}} = \frac{1}{4}$