# How do you find the limit of (sqrt (1+9x)- sqrt (1-8x))/ x as x approaches 0?

Jul 27, 2015

I found: $\frac{17}{2}$

#### Explanation:

If you try directly you get $\frac{0}{0}$:

Jul 27, 2015

"Rationalize" the numerator.

#### Explanation:

This is a trick (technique, method) that is very useful in calculus.

If we multiply $\left(\sqrt{a} - \sqrt{b}\right) \left(\sqrt{a} + \sqrt{b}\right)$, we get rid of the square roots and can try to simplify.

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

In this case:

$\frac{\sqrt{1 + 9 x} - \sqrt{1 - 8 x}}{x}$

We will multiply by $1$ in the form $\frac{\sqrt{1 + 9 x} + \sqrt{1 - 8 x}}{\sqrt{1 + 9 x} + \sqrt{1 - 8 x}}$

We get:

$\frac{\left(\sqrt{1 + 9 x} - \sqrt{1 - 8 x}\right)}{x} \cdot \frac{\left(\sqrt{1 + 9 x} + \sqrt{1 - 8 x}\right)}{\left(\sqrt{1 + 9 x} + \sqrt{1 - 8 x}\right)}$

$= \frac{\left(1 + 9 x\right) - \left(1 - 8 x\right)}{x \left(\sqrt{1 + 9 x} + \sqrt{1 - 8 x}\right)}$

$= \frac{17 x}{x \left(\sqrt{1 + 9 x} + \sqrt{1 - 8 x}\right)}$

$= \frac{17}{\sqrt{1 + 9 x} + \sqrt{1 - 8 x}}$

Now we can evaluate the limit by substitution.

${\lim}_{x \rightarrow 0} \frac{\sqrt{1 + 9 x} - \sqrt{1 - 8 x}}{x} = {\lim}_{x \rightarrow 0} \frac{17}{\sqrt{1 + 9 x} + \sqrt{1 - 8 x}}$

$= \frac{17}{\sqrt{1 + 9 \left(0\right)} + \sqrt{1 - 8 \left(0\right)}}$

$= \frac{17}{\sqrt{1} + \sqrt{1}} = \frac{17}{2}$

In a way, we have 'traded' a subtraction in the numerator that goes to $0$ for an addition in the denominator that does not go to $0$.