# How do you find the limit (sqrt(x+1)-1)/(sqrt(x+4)-2) as x->0?

Nov 15, 2016

Change the way the ratio is written.

#### Explanation:

By the time this problem is assigned, I assume students have seen things like

${\lim}_{x \rightarrow 0} \frac{\sqrt{9 + x} - 3}{x}$ which is found by 'rationalizing' the numerator:

${\lim}_{x \rightarrow 0} \frac{\left(\sqrt{9 + x} - 3\right)}{x} \cdot \frac{\left(\sqrt{9 + x} + 3\right)}{\left(\sqrt{9 + x} + 3\right)} = {\lim}_{x \rightarrow 0} \frac{1}{\left(\sqrt{9 + x} - 3\right)} = \frac{1}{6}$

In this problem, use the same trick on both the numerator and denominator

$\frac{\sqrt{x + 1} - 1}{\sqrt{x + 4} - 2} = \frac{\left(\sqrt{x + 1} - 1\right)}{\left(\sqrt{x + 4} - 2\right)} \cdot \frac{\left(\sqrt{x + 1} + 1\right) \left(\sqrt{x + 4} + 2\right)}{\left(\sqrt{x + 1} + 1\right) \left(\sqrt{x + 4} + 2\right)}$

$= \frac{\left(x + 1 - 1\right) \left(\sqrt{x + 4} + 2\right)}{\left(x + 4 - 4\right) \left(\sqrt{x + 1} + 1\right)}$

$= \frac{x \left(\sqrt{x + 4} + 2\right)}{x \left(\sqrt{x + 1} + 1\right)}$

$= \frac{\sqrt{x + 4} + 2}{\sqrt{x + 1} + 1}$ $\text{ }$ (for $x \ne 0$)

${\lim}_{x \rightarrow 0} \frac{\sqrt{x + 1} - 1}{\sqrt{x + 4} - 2} = {\lim}_{x \rightarrow 0} \frac{\sqrt{x + 4} + 2}{\sqrt{x + 1} + 1}$

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