To evaluate: $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$
Formula used:
L'Hospital's rule
Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where
$\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$ or $\pm \infty$
then $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$
As $x \rightarrow 0$,
we have $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{0}{0}$
This represents an indeterminate form. Thus applying L'Hospital's rule,
we get $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}\left(x^2-4\right)}{\frac{d}{d x}(\sqrt{x+2}-\sqrt{3 x-2})}$
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\lim _{x \rightarrow 2} \frac{2 x}{\frac{1}{2 \sqrt{x+2}}-\frac{3}{2 \sqrt{x-2}}}$
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{4}{\frac{1}{2 \sqrt{2+2}}-\frac{3}{2 \sqrt{5-2}}}$
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{8}{\frac{1}{2}-\frac{3}{2}}$
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=-8$
Thus, the value of $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$ is $-8$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.