Question

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

Answer 1

We have,

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

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

So, apply L-Hospital rule

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

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

$\Rightarrow \frac{\frac{3}{2\sqrt{9}}}{\frac{1}{\sqrt{6-4}}}$

$\Rightarrow \frac{\sqrt{2}}{2}$

$\Rightarrow \frac{1}{\sqrt{2}}$

Hence, this is the answer.