Question

Evaluate the limit:

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

Answer 1

At $x\to 1,\frac{x-1}{\sqrt{x^2+3}-2}\text{becomes}\frac{0}{0}form$

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

On rationalising denominator, we get

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

$=\underset{x\to 1}{lim}\frac{(x-1)(\sqrt{x^2+3}+2)}{({\left(\sqrt{x^2+3}\right)}^2-2^2)}$

$[\because (a-b\left)\right(a+b)=(a^2-b^2\left)\right]$

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

$=\underset{x\to 1}{lim}\frac{(x-1)(\sqrt{x^2+3}+2)}{\left(x^2-1\right)}$

$=\underset{x\to 1}{lim}\frac{(x-1)(\sqrt{x^2+3}+2)}{(x-1)(x+1)}\left[\right(x-1)\ne 0]$

$=\underset{x\to 1}{lim}\frac{(\sqrt{x^2+3}+2)}{(x+1)}$

$=\frac{(\sqrt{1^2+3}+2)}{(1+1)}$

$=2$