Question

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

Answer 1

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


It is of the form $\frac{0}{0}$.
Rationalising the denominator:

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

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

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

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

=$\frac{\sqrt{1+3}+2}{1+1}$

= $\frac{4}{2}$

= 2