Question

${lim}_{x\to 3}\frac{\sqrt{4-x}-\sqrt{x-2}}{15-5x}$

Answer 1

We have,

${lim}_{x\to 3}\left(\frac{\sqrt{4-x}-\sqrt{x-2}}{15-5x}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

$\Rightarrow {lim}_{x\to 3}\left(\frac{\frac{1}{2\sqrt{4-x}}\times (0-1)-\frac{1}{2\sqrt{x-2}}}{0-5}\right)$

$\Rightarrow {lim}_{x\to 3}\left(\frac{\frac{1}{2\sqrt{4-x}}\times (0-1)-\frac{1}{2\sqrt{x-2}}}{0-5}\right)$

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

$\Rightarrow \frac{\frac{1}{2\sqrt{4-3}}+\frac{1}{2\sqrt{3-2}}}{5}$

$\Rightarrow \frac{\frac{1}{2}+\frac{1}{2}}{5}$

$\Rightarrow \frac{1}{5}$

Hence, this is the answer.