<s>[INST] 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?
 [/INST] Yes indeed, that's the way to go about it. Now, we have $$\begin{align} \lim_{x \to 2}\frac{x-2}{\sqrt{x^2+5}-3} & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{x^2 - 4} \\ \\ & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{(x - 2)(x+2)} \\ \\ & \overset{x\neq 2}{=} \lim_{x \to 2}\dfrac {\sqrt{x^2 + 5} + 3}{x+2} \\ \\ 
& = \dfrac 64 = \frac 32\end{align}$$
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