How to find the value of $\ \lim_{x\to 3^+}\frac{\sqrt{x^2-9}}{x-3}$? How do I find the value of this limit?
$$\lim_{x\to 3^+}\frac{\sqrt{x^2-9}}{x-3}$$
It says that it's approaching from right side to 3 right?
I tried subsitituting the 3 into the variables, and got 0, and the answer says that it's positive infinity.
I tried using those graphing software, I don't know how it's positive infinity.
I'm pretty sure there's something I'm doing wrong.
 A: We have
\begin{align}
\lim_{x\rightarrow 3^{+}}\frac{\sqrt{x^2-9}}{x-3}& =\lim_{x\rightarrow 3^{+}}\frac{\sqrt{\left(x+3\right)\left(x-3\right)}}{x-3}\tag{1} \\[1ex]
& = \lim_{x\rightarrow 3^{+}}\frac{\sqrt{\left(x+3\right)\left(x-3\right)}}{\sqrt{\left(x-3\right)^2}} \tag{2}\\[1ex]
& =\lim_{x\rightarrow 3^{+}}\sqrt{\frac{\left(x+3\right)}{\left(x-3\right)}} \tag{3}\\[1ex]
& = \lim_{x\rightarrow 3^{+}}\sqrt{1+\frac{6}{x-3}}\tag{4} \\[1ex]
& =\infty.\tag{5}
\end{align}
Here is a graph as well with gridlines at $x=-3,3$:

Notice that the function is undefined on $\mathbb{R}^2$ between $x\in\left[-3,3\right)$.
A: You should know, that
$$ \left(\forall a \in \mathbb{R}^{+}\right)(\lim_{x\to 0+}\frac{a}{x} = +\infty)$$

Hint:
$$\lim_{x\to3+}x-3 = \lim_{x\to0+}x$$
Carefully! You cannot directly considering! You haven't some number in nominator. You receive $\frac{0}{0}$ — it is indeterminate form.
Hint2: Shortcut formulas
$$ (x^2-9) = (x+3)(x-3)$$
Hint3:
$$ x-3 \geq 0 \Longrightarrow x-3 = \sqrt{(x-3)^2}$$
A: Note that for $x > 3$, we have
$$\dfrac{\sqrt{x^2-9}}{x-3} = \sqrt{\dfrac{x+3}{x-3}}$$
Now look at what happens to the numerator and denominator as $x \to 3^+$.
A: $$\lim_{x\to 3^+} \dfrac{\sqrt{x^2-9}}{x-3} = \lim_{x\to 3^+} \sqrt{\dfrac{x+3}{x-3}} 
= \lim_{x\to 3^+} \sqrt{\dfrac{6}{x-3}}  = +\infty
$$
A: Its clearly +∞ try substituting x as 3+h where h tends to zero simplify and get the answer.. 
