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.
On
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 $x \to 3^+$.
On
$$\lim_{x\to 3^+} \dfrac{\sqrt{x^2-9}}{x-3} = \lim_{x\to 3^+} \sqrt{\dfrac{x+3}{x-3}} = \lim_{x\to 3^+} \sqrt{\dfrac{6}{x-3}} = +\infty $$
On
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:
$$ x-3 \geq 0 \Longrightarrow x-3 = \sqrt{(x-3)^2}$$
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 $x\in\left[-3,3\right)$.