How can I prove a limit is infinity?

Question

How can I prove that the $\lim \limits_{x \to 1^+} \frac{x^2}{x-1}=\infty$ using the $\epsilon -\delta$ definition of a limit? I think I start with $\forall$M>0, want $\delta$>0. After that I'm not sure.

Answer 1

You have to show: $\forall \epsilon >0 \; \exists \delta>1$ such that: $\delta>x>1 \implies$ $\frac{x^2}{x-1} > \epsilon$.

Note that the function is non-increasing (monotonous) on $[1,2]$. Therefore it is sufficient to just ask for a $1 < \delta <2$ with the property that: $\frac{\delta^2}{\delta-1} > \epsilon$. The condition above is then automatically true for any $1 < x < \delta$.

$ \delta = \frac{1}{2} (\epsilon+\sqrt{\epsilon-4} \sqrt{\epsilon})$ satisfies this condition.

Answer 2

You need to show that

$$\forall M \in \mathbb{R} \exists \delta s.t. \delta > x > 1 \implies \frac{x^2}{x-1} > M$$

Answer 3

With $x=1+\delta$ for $\delta>0$, we can write $$ \frac{x^2}{x-1}=x+1+\frac{1}{x-1}=2+\delta+\frac{1}{\delta} $$ so for any $M>0$, we only need to pick $\delta\in(0,\frac{1}{M})$ (say, $\frac{1}{2M}$) to ensure that the RHS above is greater than $M$.