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?
 A: From what you have got, you can simply cancel the factor $(x-2)$ from both numerator and denominator. After that, you can substitute $x=2$ into the expression and obtain the limit.
A: Yes indeed, that's the way to go about it. Now, we have $$\begin{align} \lim_{x \to 2}\frac{x-2}{\sqrt{x^2+5}-3} & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{x^2 - 4} \\ \\ & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{(x - 2)(x+2)} \\ \\ & \overset{x\neq 2}{=} \lim_{x \to 2}\dfrac {\sqrt{x^2 + 5} + 3}{x+2} \\ \\ 
& = \dfrac 64 = \frac 32\end{align}$$
A: You are on the right track.
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}=\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3} \frac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3}=\lim_{x\to 2} \frac{(x-2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)}$ simplifying
$$\lim_{x\to 2} \frac{(\sqrt{x^2+5}+3)}{(x+2)}$$ and evaluating at $2$ we get
$$\lim_{x\to 2} \frac{(\sqrt{x^2+5}+3)}{(x+2)}= \frac{(\sqrt{4+5}+3)}{(2+2)}=\frac{6}{4}=\frac{3}{2}$$
A: 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}
$
