Evaluate the limit $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$ I need to evaluate the following limit:
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?

In
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
let $x = y+2$.
This becomes
$\begin{align}
\lim_{y\to 0} \frac{y}{\sqrt{(y+2)^2+5}-3}
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+4+5}-3}\\
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+9}-3}\\
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+9}-3}\frac{\sqrt{y^2+4y+9}+3}{\sqrt{y^2+4y+9}+3}\\
&=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{(\sqrt{y^2+4y+9}-3)(\sqrt{y^2+4y+9}+3)}\\
&=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{(y^2+4y+9)-9}\\
&=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{y^2+4y}\\
&=\lim_{y\to 0} \frac{\sqrt{y^2+4y+9}+3}{y+4}\\
&=\frac{\sqrt{9}+3}{4}\\
&=\frac{6}{4}\\
&=\frac{3}{2}\\
\end{align}
$
