Two Limit questions

Question

$$\lim_\limits{x\to 4}\;\frac{x^2 - 16}{3x - 12} $$

Factoring doesn't help so I'm not sure what to do

Also,

$$\lim_\limits{ x\to -\infty}\;\frac{x^3 - 5x^2}{3x}$$

If you put a really big negative number in $x^3$ it will still be negative, so I'm not sure why the answer is $\infty$ and not $-\infty $???

Answer 1

$$\lim_{x\rightarrow4}\frac{x^2-16}{3x-12}=\lim_{x\rightarrow4}\frac{(x-4)(x+4)}{3(x-4)}=\lim_{x\rightarrow4}\frac{x+4}{3}=\frac{8}{3}.$$ Actually, if $x\rightarrow4$ then by the definition of the limit $x\neq4$.

$$\lim_{x\rightarrow-\infty}\frac{x^3-5x^2}{3x}=\frac{1}{3}\lim_{x\rightarrow-\infty}x^2\left(1-\frac{5}{x}\right)=+\infty.$$

Answer 2

$$\lim_{x\to-\infty}\frac{x^3-5x^2}{3x}=\lim_{x\to-\infty}\frac{x^2-5x}{3}$$ since $x^2>0$ and $x^2$ grows more rapidly than $5x$, $$\lim_{x\to-\infty}\frac{x^2-5x}{3}=\infty$$