How do I prove the value of $\lim_{x \to 3} {{x^3 - 27} \over {x - 3}}$ is 27? Find the value of $\lim_{x \to 3} {{x^3 - 27} \over {x - 3}}$ and use an $\epsilon-\delta$ proof to show your answer is correct.
This is a review problem, I always forget how to do this. All I need is to see how it's done. 
Could someone show me a detailed proof please? Thank you!
 A: $$\lim_{x \to 3}\frac{x^3-27}{x-3}=\lim_{x \to 3}\frac{(x-3)(x^2+3x+9)}{x-3}=\lim_{x \to 3}x^2+3x+9$$
So Proceed by saying the usual:
$$\forall \quad \epsilon>0 \quad \exists \quad \delta>0 : |x-3|<\delta \Longrightarrow |\frac{x^3-27}{x-3}-27|<\epsilon$$
Replace the first form with
$$\forall \quad \epsilon>0 \quad \exists \quad \delta>0 : |x-3|<\delta \Longrightarrow |x^2+3x+9-27|<\epsilon$$
You can simplify the epsilon modulus with
$$|x^2+3x-18|<\epsilon$$
So you want to end up at 
$$|x-3||x+6|<\epsilon$$
When this happens, it helps if you define your delta (without loss of generality). e.g. $$Let \quad \delta :=1$$
Open up this modulus involving delta:
$$if \quad |x-3| < \delta=1$$
$$\Longrightarrow -1<x-3<1$$
$$\Longrightarrow-(9+1)<9-1<x-3+9=x+6<9+1$$
Ie
$$\Longrightarrow |x+6|<10$$
so you find a condition for your $\epsilon$
So we require $$\epsilon := 10\delta $$
(since $\delta$ equaled 1 for x+3 and we ended up with 10!)
and then just work in reverse!
A: Let $\epsilon>0$. 
Then,
\begin{align}
|&\dfrac{x^3-27}{x-3}-27|
\\=&|(x^2+3x+9)-27|
\\=&|x^2+3x-18|
\\=&|((x-3)+3)^2+3((x-3)+3)-18|
\\=&|(x-3)^2+5(x-3)|
\\\leq& |x-3|^2+5|x-3|
\end{align}
Can you proceed from here?
