<s>[INST] How would I evaluate this limit? $\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$ How would I evaluate the following limit by hand?
$\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$
Thanks in advance.
 [/INST] $$\begin{align} \lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}} & = \lim_{x \to 3+}\; \frac{\left(\sqrt{x - 3}\;\right)^2}{\sqrt{(x - 3)(x+3)}}\\ \\ 
& = \lim_{x \to 3+}\;\frac{\left(\sqrt{x-3}\;\right)^2}{(\sqrt{x-3}\;)(\sqrt{x+3}\;)}\\ \\
& = \lim_{x \to 3+}\;\frac{\sqrt{x - 3}}{\sqrt{x+3}}\\ \\
& = \frac 06\; = \;0\end{align}$$
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