How can $\lim_{x\to3}\left(\frac{\sqrt[3]{32x-96}}{x^{2}-2x-3}\right)$ be shown to equal $2$?

Question

The question given in my textbook is "Compute the limit $\lim\limits_{x\to3}\left(\frac{\sqrt[3]{32x-96}}{x^{2}-2x-3}\right)$." The solution given is merely the answer $2$, but I'm not sure how this is obtained, especially after looking at a graph. How can this be shown to equal $2$? This question is in the context of the introductory chapter to limits, so some of the more advanced tools are not (I imagine) supposed to be used. Thank you in advance for any help!

Edit: Appreciate the help from everyone, it is obvious now that the book didn't format the root correctly. Also thank you to the user that fixed my Latex code.

Answer 1

Factor the quantity inside the cube root, and factor the denominator: $$\frac{\sqrt[3]{32x-96}}{x^2-2x-3} = \frac{\sqrt[3]{32(x-3)}}{(x+1)(x-3)} = \frac{\sqrt[3]{32}}{(x+1)(x-3)^{\frac23}}$$ Take the limit as $x\to 3$, and we get $\frac{\sqrt[3]{32}}{4}=\frac1{\sqrt[3]{2}}$ times a $\frac1{0}$ form. The quantity $(x-3)^{\frac23}$ we're dividing by is positive on both sides, so the limit is $\infty$. Uh, oops. There's a mistake here, and it looks like it's the textbook's statement of the problem.

Answer 2

Either you or the book made a mistake. The limit should probably be:

$$ \lim_{x \to 3} \sqrt[3]{\frac{32x-96}{x^2 - 2x - 3}} $$

Answer 3

As JavaMan correctly indicates, this should be: $$\lim_{x\to 3} \bigg(\frac{32x-96}{x^2-2x-3}\bigg)^\frac13$$

We use two properties. Firstly that when $B(x)=A(x)\cdot \frac{x-t}{x-t}$ for any function $A(x)$ is that function with a discontinuity at $x=t$, and it holds that $$\lim_{x\to t} B(x)=A(t)$$

The second property is the algebra of limits. Namely; $$\lim_{x\to a}(B(x))^K =\bigg(\lim_{x\to a} B(x) \bigg)^K$$

Use $B(x)=\frac{32x-96}{x^2-2x-3}=\frac{32(x-3)}{(x+1)(x-3)}$ then $K=\frac 13$ to see what I'm driving at.