# How do you find the limit of (x-4)/(sqrt(x-3)-sqrt(5-x)) as x approaches 4?

Apr 13, 2015

The answer is: $1$.

This limit shows up in the $\frac{0}{0}$ indecision form, so we have to rationalise.

${\lim}_{x \rightarrow 4} \frac{x - 4}{\sqrt{x - 3} - \sqrt{5 - x}} =$

$= {\lim}_{x \rightarrow 4} \left[\frac{x - 4}{\sqrt{x - 3} - \sqrt{5 - x}} \cdot \frac{\sqrt{x - 3} + \sqrt{5 - x}}{\sqrt{x - 3} + \sqrt{5 - x}}\right] =$

$= {\lim}_{x \rightarrow 4} \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{x - 3 - 5 + x} =$

$= {\lim}_{x \rightarrow 4} \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2 x - 8} =$

$= {\lim}_{x \rightarrow 4} \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2 \left(x - 4\right)} =$

$= {\lim}_{x \rightarrow 4} \frac{\left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2} = \frac{\sqrt{4 - 3} + \sqrt{5 - 4}}{2} = 1$.

Apr 13, 2015

Rationalize the denominator by multiplying the fraction by $1$ in the form:

$\frac{\sqrt{x - 3} + \sqrt{5 - x}}{\sqrt{x - 3} + \sqrt{5 - x}}$

This works because $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$,

so $\left(\sqrt{a} - \sqrt{b}\right) \left(\sqrt{a} + \sqrt{b}\right) = a - b$. It looks like this:

$\frac{\left(x - 4\right)}{\left(\sqrt{x - 3} - \sqrt{5 - x}\right)} \frac{\left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{\left(\sqrt{x - 3} + \sqrt{5 - x}\right)}$

 =((x-4) (sqrt(x-3)+sqrt(5-x)))/((x-3)-(5-x)

$= \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2 x - 8}$

$= \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2 \left(x - 4\right)}$

$= \frac{\sqrt{x - 3} + \sqrt{5 - x}}{2}$

So, we have:

${\lim}_{x \rightarrow 4} \frac{x - 4}{\sqrt{x - 3} - \sqrt{5 - x}} = {\lim}_{x \rightarrow 4} \frac{\sqrt{x - 3} + \sqrt{5 - x}}{2}$

$= \frac{2}{2} = 1$