how to define the functions $f(x)$ and $g(x)$ by a given equation

Question

Define functions $f ( x )$ and $g ( x )$ by $f ( x ) = \frac { \sqrt [ 3 ] { x } - 2 } { x - 8 } , \quad g ( x ) = \frac { 1 } { \sqrt [ 3 ] { x } ^ { 2 } + 2 \sqrt [ 3 ] { x } + 4 }$

(a) By multiplying $f ( x )$ by $$\frac { \sqrt [ 3 ] { x } ^ { 2 } + 2 \sqrt [ 3 ] { x } + 4 } { \sqrt [ 3 ] { x } ^ { 2 } + 2 \sqrt [ 3 ] { x } + 4 }$$ show that $f ( x ) = g ( x )$ for all real numbers $x \neq 2$

(b) Using the Equal Limit Theorem from lectures, find $$\lim _ { x \rightarrow 8 } \frac { \sqrt [ 3 ] { x } - 2 } { x - 8 }$$

The question is in the link provided. For (a), I solved $f(x)=g(x)$ by using a value that gives a whole number when cube rooted. So I subbed in 64 for x because cube root 64 is 4. After solving, I got $\frac{1}{28}$ for both $f(x)$ and $g(x)$. But I'm not sure how to solve (b) using the equal limit theorem. Any help is appreciated.

Answer 1

For part b)

I looked up "equal limit theorem" and didn't really find anything on this.

If you try and evaluate the limit algebraically it doesn't really work out, however you can use L Hopitals rule on it.

The limit ends up being 1/4

Answer 2

For a you need to show it holds for all $x$ except $2$, not for one specific $x$. It sounds like you just showed it for $x=64$. You are expected to carry out the multiplication suggested and manipulate it to get $g(x)$.

For b the point is that $g(x)$ does not have an indeterminate $\frac 00$ at $x=8$. As it is continuous you can just substitute in $x=8$ to find the limit.

The fraction you are supposed to multiply by was found by looking at $x-8$ as $(\sqrt[3]x)^3-8$, which is a difference of cubes and using the standard factorization.