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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.

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

1 Answer 1

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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 $\lvert f(x)\rvert <M$ 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.

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