Let $f:\mathbb{R}\to\mathbb{R}$ be a function. Prove that if $\lim_{x\to\infty}f(x) = \infty$, then $\lim_{x\to\infty}\frac{1}{f(x)} = 0$.
I'm only part of the way there. I started with a version of the definition of an infinite limit as x approaches infinity:$$\lim_{x\to\infty}f(x)=\infty \iff \forall\space M>0\space\exists\space N>0:x>N\implies f(x)>M$$
And if the limit of $1/f(x)$ is $0$, then by def: $$\forall\space\epsilon>0,\exists\space\delta>0:x>\delta\implies\left|\frac{1}{f(x)}\right|<\epsilon$$
I'm unsure as to how to marry these conditions together without loss of generality. I was going to choose $\delta = N$ and $\epsilon = \frac{1}{M}$ but this feels wrong. I'm not sure if I'm even allowed just choose. Any help would be much appreciated.
Best Answer
You have it exactly right.
I think of $\epsilon-\delta$ proofs as challenge and response.
Your goal is to show that $\lim_{x\to\infty} \frac{1}{f(x)} = 0$.
To do so you must be able to respond to the challenge: "Here is $\epsilon$" with the response: "This is a $\delta$ such that if $x > \delta$ then $|1/f(x)| < \epsilon$."
So let's get to it. You are given $\epsilon$. You want $|1/f| < \epsilon$, which you can get if $f > 1/\epsilon$. Since $f\to\infty$, there is some $N$ such that $x > N$ means that $f > 1/\epsilon$. And now you have a response to the challenge!
"Here is $\epsilon$."
"Here is $\delta = N$, so that if $x > \delta$ then $f(x) > 1/\epsilon$, which means $1/f(x) < \epsilon$."