determine if the statement with limit is true

Question

I want to determine if the statement $$\lim_{x\to\ a} f(x) = \infty \Rightarrow \lim_{x\to\ a}\frac{1}{f(x)} = 0$$

is true or not (by proving it or proving a contradiction).

I know that I have a definition of limits that could be of help. That is, if the function has a limit $A$ when $x \rightarrow a$ then there is a number $ε > 0$ and $w$ such that

$$|x-a|< w \Rightarrow |f(x)-A| < ε . $$

And I have all the other rules for limits. But I don't know how to go about this problem.

Answer 1

HINT

We have to show that for each $\epsilon > 0 \ \exists \delta > 0$ such that whenever $|x-a| < \delta$ you have $$ \epsilon > \left|\frac{1}{f(x)} - 0 \right| = \left|\frac1{f(x)}\right| \iff |f(x)| > 1/\epsilon. $$

Assume that as $x \to a$, we have $f(x) \to \infty$, in other words, that $\forall N > 0 \ \exists \delta>0$ such that $f(x) > N$ whenever $|x - a| < \delta$.

Can you see how to pick $\epsilon = \epsilon(N)$ to make what you want happen?

Answer 2

We say that $\lim_{x\to\ a} f(x) = \infty$, if $\forall M>0, \exists δ>0$ s.t. whenever $\vert x-a \vert < δ$, then $f(x) > M $.

We say that $\lim_{x\to\ a}\frac{1}{f(x)} = 0$, if $\forall ε>0, \exists δ>0$ s.t. whenever $\vert x-a \vert < δ$, then $\left|\frac{1}{f(x)} - 0 \right| <\epsilon$.

It suffices to choose $\epsilon = \frac{1}{M}$, in which case $\left|\frac{1}{f(x)} - 0 \right| <\epsilon$ holds.

Answer 3

Alternatively we may use Heine's definition of the limit of a sequence.
Let $x_n$ be a sequence so that $x_n \rightarrow a => \lim_{n\to \infty} f(x_n)=\infty$. We have: $\lim_{x\to a} \frac{1}{f(x)}=\lim_{n\to \infty} \frac{1}{f(x_n)}=\frac{1}{\lim_{n\to\infty} f(x_n)}=\frac{1}{\infty}=0$