Given  (sqrt(x+1) - sqrt(2x+1)) / (sqrt(3x+4) - sqrt(2x+4) ]  how do you find the limit as x approaches 0?

Nov 22, 2016

Change the way it is written to avoid $\frac{0}{0}$.

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, we have to try something, so late's use the same trick on both the numerator and denominator

(sqrt(x+1)-sqrt(2x+1))/(sqrt(3x+4)-sqrt(2x+4)) = ((sqrt(x+1)-sqrt(2x+1)))/((sqrt(3x+4)-sqrt(2x+4))) * ((sqrt(x+1)+sqrt(2x+1))(sqrt(3x+4)+sqrt(2x+4)))/((sqrt(x+1)+sqrt(2x+1))(sqrt(3x+4)+sqrt(2x+4))

$= \frac{\left(x + 1 - \left(2 x + 1\right)\right) \left(\sqrt{3 x + 4} + \sqrt{2 x + 4}\right)}{\left(x + 4 - 4\right) \left(\sqrt{x + 1} + \sqrt{2 x + 1}\right)}$

$= \frac{- x \left(\sqrt{3 x + 4} + \sqrt{2 x + 4}\right)}{x \left(\sqrt{x + 1} + \sqrt{2 x + 1}\right)}$

$= \frac{- \left(\sqrt{3 x + 4} + \sqrt{2 x + 4}\right)}{\sqrt{x + 1} + \sqrt{2 x + 1}}$ $\text{ }$ (for $x \ne 0$)

${\lim}_{x \rightarrow 0} \frac{\sqrt{x + 1} - \sqrt{2 x + 1}}{\sqrt{3 x + 4} - \sqrt{2 x + 4}} = {\lim}_{x \rightarrow 0} \frac{- \left(\sqrt{3 x + 4} + \sqrt{2 x + 4}\right)}{\sqrt{x + 1} + \sqrt{2 x + 1}}$

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

The same algebra with simplified notation gets us

$\frac{\sqrt{a} - \sqrt{b}}{\sqrt{c} - \sqrt{d}} = \frac{\left(\sqrt{a} - \sqrt{b}\right)}{\left(\sqrt{c} - \sqrt{d}\right)} \cdot \frac{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{c} + \sqrt{d}\right)}{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{c} + \sqrt{d}\right)}$

$= \frac{\left(a - b\right) \left(\sqrt{c} + \sqrt{d}\right)}{\left(c - d\right) \left(\sqrt{a} + \sqrt{b}\right)}$