Prove by definition $$\lim_{x \to 2} \frac{1}{(x-2)^3}=\infty$$
How should I approach this, if the limit is $\infty$?
2026-04-03 05:14:32.1775193272
Proof by definition limit equals to infinity
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For $$\lim_{x\to 2^+}\frac{1}{(x-2)^3}=+\infty, $$ if $0<x-2<1$, then $$(x-2)^3\leq (x-2)\implies \frac{1}{(x-2)^3}\geq \frac{1}{x-2}.$$ Then, if $M>0$, set $\delta=\min\{\frac{1}{M},1\}$, to get $$\frac{1}{(x-2)^3}>M$$ when $0<x-2<\delta$ what prove the claim. I let you do the case $$\lim_{x\to 2^-}\frac{1}{(x-2)^3}=-\infty .$$