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

Mar 28, 2016

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#### Explanation:

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

${\lim}_{x \rightarrow 0} \sqrt{1 + 2 x} = 1$ and

${\lim}_{x \rightarrow 0} \frac{\sqrt{1 - 4 x}}{x}$ does not exist, because as $x \rightarrow {0}^{+}$, we have $\frac{\sqrt{1 - 4 x}}{x}$ increases without bound.

Therefore, the difference, ${\lim}_{x \rightarrow 0} \left(\sqrt{1 + 2 x} - \frac{\sqrt{1 - 4 x}}{x}\right)$ does not exist because as $x \rightarrow {0}^{+}$, the value of the expression decreases without bound.

If you want ${\lim}_{x \rightarrow 0} \frac{\sqrt{1 + 2 x} - \sqrt{1 - 4 x}}{x}$ then "rationalize" the numerator.

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

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

$= \frac{6 x}{x \left(\sqrt{1 + 2 x} + \sqrt{1 - 4 x}\right)}$

$= \frac{6}{\sqrt{1 + 2 x} + \sqrt{1 - 4 x}}$

Now we have

${\lim}_{x \rightarrow 0} \frac{\sqrt{1 + 2 x} - \sqrt{1 - 4 x}}{x} = {\lim}_{x \rightarrow 0} \frac{6}{\sqrt{1 + 2 x} + \sqrt{1 - 4 x}}$

$> \frac{6}{\sqrt{1} + \sqrt{1}} = 3$