How would I evaluate this limit? $\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$

Question

How would I evaluate the following limit by hand?

$\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$

Thanks in advance.

Answer 1

$$\begin{align} \lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}} & = \lim_{x \to 3+}\; \frac{\left(\sqrt{x - 3}\;\right)^2}{\sqrt{(x - 3)(x+3)}}\\ \\ & = \lim_{x \to 3+}\;\frac{\left(\sqrt{x-3}\;\right)^2}{(\sqrt{x-3}\;)(\sqrt{x+3}\;)}\\ \\ & = \lim_{x \to 3+}\;\frac{\sqrt{x - 3}}{\sqrt{x+3}}\\ \\ & = \frac 06\; = \;0\end{align}$$

Answer 2

Hint: $$ \lim_{(x\to 3^+)}{\frac{(x-3)}{\sqrt{(x^2-9)}}} = \lim_{(x\to 3^+)}{\sqrt \frac{(x-3)^2}{(x-3)(x+3)}} $$

$$ \lim_{(x\to 3^+)}{\sqrt\frac{(x-3)}{(x+3)}} = \frac {3-3}{3+3} = \frac{0}{6} = 0 $$