Prove that the limit $\lim_{x \to 2}\frac{1}{(x-2)^3}$ is infinity using $\epsilon$-$\delta$

Question

Prove that $\lim_{x \to 2}\frac{1}{(x-2)^3}=\infty$

Am I allowed to replace this problem with $\lim_{x \to 2}{(x-2)^3}=\frac{1}{\infty}(=0)$

And therefore for $0<|x-2|<\delta$ there is $\delta^3<\epsilon$ so will take $\delta=\frac{\epsilon}{2}$?

Answer 1

For every $M > 0$, if $0 < |x-2| < 1/M^{1/3}$, then $|x-2|^{-3} > M$.

Loosely speaking: if $1/\infty = 0$, then $1 = 0\cdot \infty = 0$; hence division by $\infty$ is attached no meaning.

Answer 2

First note that the function $f(x)=\dfrac{1}{(x-2)^3}$ change sign at $x=2$ so we can suppose that the limit fot $x>2$ and the limit for $x<2$ are different.

For $\lim_{x \to 2^+} f(x)$ chose $N>0$ and consider the system: $$ \begin{cases} x>2\\ \dfrac{1}{(x-2)^3}>N \end{cases} $$ solving we find: $$\begin{cases} x>2\\ x<\sqrt[3]{\dfrac{1}{N}}+2 \end{cases}$$ so the solution is $2<x<\sqrt[3]{\dfrac{1}{N}}+2$. This means that, choosing $\epsilon=\sqrt[3]{\dfrac{1}{N}}$ we have that:

for every $N>0$, there exists $\epsilon>0$ such that $f(x)>N$ if $2<x <2+\epsilon$ .

And this is the definition of an infinite limit form right.

For the limit from left you can do the same, with some attention to the signs.