Question

Evaluate : $\underset{x\to 3}{lim}\frac{x^2-4x+3}{\sqrt{2x+3}-3}$

Answer 1

We have, $\underset{x\to 3}{lim}\frac{x^2-4x+3}{\sqrt{2x+3}-3}$

Let $L=\underset{x\to 3}{lim}\frac{x^2-4x+3}{\sqrt{2x+3}-3}$

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

So, apply L-Hospital rule,

$L=\underset{x\to 3}{lim}\frac{2x-4+0}{\frac{1}{2\sqrt{2x+3}}\times 2-0}$

$=\underset{x\to 3}{lim}\frac{2x-4}{\frac{1}{\sqrt{2x+3}}}$

$=\underset{x\to 3}{lim}(2x-4)\sqrt{2x+3}$

$=(2\times 3-4)\sqrt{2\times 3+3}$

$=(6-4)\sqrt{6+3}$

$=\left(2\right)\sqrt{9}$

$L=\left(2\right)\times 3=6$

Hence, this is the answer.