Proving for infinite limits when $x$ tends to $a$.

Question

I am trying to solve this: $$\lim_{x\to 2^+}\left(\frac{1}{2-x} - \frac{3}{8-x^3}\right)$$ I tried to combine the fractions and got this: $$\lim_{x\to 2^+}\frac{(x+1)^2}{8-x^3}$$ I plotted the graph and, intuitively, I think I know why it should tend towards negative infinity. But I am stuck since I do not know how to prove that a function tends to infinity. After trawling through the formal definition of limits, I think I can sort of understand a little that I need to prove that $f(x)$ is always greater than some value M which is somehow derived from the distance $x-a$. I am probably wrong with my understanding, so can someone help me understand the formal definition and how to prove that a limit is infinity please? I can't seem to understand M and N and $\epsilon$ and $\delta$. Edit: I am so sorry, I typed the limit wrong. It should be 2, not 0. Thank you very much to everyone who answered me (And guessed my real question.).

Answer 1

$$\lim_{x\to 0}\left(\frac{1}{2-x} - \frac{3}{8-x^3}\right)=\frac{1}{2}-\frac{3}{8}=\frac{1}{8}$$ Maybe $$\lim_{x\to 2^+}\left(\frac{1}{2-x} - \frac{3}{8-x^3}\right)=\lim_{x\to 2^+}\frac{(x+1)^2}{(2-x)(x^2+2x+4)}=-\infty$$ and $$\lim_{x\to 2^-}\left(\frac{1}{2-x} - \frac{3}{8-x^3}\right)=+\infty.$$

Answer 2

Notice that $8-x^3=(2-x)(4+2x+x^2)$, thus $$ \frac{1}{2-x}-\frac{1}{8-x^3}=\frac{1}{2-x}\left(1-\frac{1}{x^2+2x+4}\right) $$ $\lim\limits_{x\rightarrow 2}\left(1-\frac{1}{x^2+2x+4}\right)=1-\frac{1}{12}=\frac{11}{12}$. Thus $\lim\limits_{x\rightarrow 2^{\pm}}\left(\frac{1}{2-x}-\frac{1}{8-x^3}\right)=\mp \infty$.