# Trouble showing that $\lim\limits_{x\to \infty} f(x)=\infty \implies \lim\limits_{x\to \infty} \frac{1}{f(x)}=0$

So, here's what I've done:

$$\lim\limits_{x\to \infty} f(x)=\infty \implies \lim\limits_{x\to \infty} \frac{1}{f(x)}=0$$

By definition of the first limit: $$\forall M>0, \exists N>0:x>N \implies \lvert{f(x)}\rvert>M$$

And for the second limit: $$\forall \epsilon>0, \exists N'>0:x>N' \implies \frac{1}{\lvert{f(x)}\rvert}<\epsilon$$

We need $$\frac{1}{\lvert{f(x)}\rvert}\lt \epsilon$$ , so $$f(x)\gt \frac{1}{\epsilon}$$, but we know that $$f(x)\rightarrow \infty$$, wich implies that there exists $$N\gt 0:x>N \implies f(x)>\frac{1}{\epsilon}$$

After that I tried to attach $$N'$$ to $$f(x)>\frac{1}{\epsilon}$$. But isn't that the $$M$$ of the definition? I still don't know how to apply the theory well. Any advise is appreciated.

• In the third line you need $f(x) >M$ and this is equivalent to $f(x) >0$ and $\frac 1 {f(x)} <\frac 1 M$. Commented Sep 4, 2023 at 0:16
• Mistake: $|f(x)| <M$. It should have been $|f(x)| >M$ as per the definition if $\lim\limits_{n\to \infty} f(x)$ Commented Sep 4, 2023 at 0:17
• Just take $M = \frac{1}{\varepsilon}$. and apply the first limit to obtain the second. Commented Sep 4, 2023 at 0:20

The first definition should actually be that for all $$M>0$$, there exists an $$x_0\in\mathbb{R}$$ such that

$$f(x)>M$$

for all $$x\geq x_0$$. Notice how what you wrote was $$\lvert f(x)\rvert when it should be $$f(x)>M$$.

Now let $$\varepsilon>0$$, and choose $$x_0\in\mathbb{R}$$ such that

$$f(x)>\frac{1}{\varepsilon}$$

for all $$x\geq x_0$$ (this is just the definition from above where we take $$M=\frac{1}{\varepsilon}$$). Rearranging we get that

$$\frac{1}{f(x)}<\varepsilon$$

for all $$x\geq x_0$$. Finally, as $$\frac{1}{\varepsilon}>0$$, we have that $$f(x)>0$$ for $$x\geq x_0$$, i.e. $$f(x)=\lvert f(x)\rvert$$. Consequently

$$\left\lvert\frac{1}{f(x)}\right\rvert<\varepsilon$$

for all $$x\geq x_0$$. The result follows.