Is this true that if limit approaches infinity the function equals to zero?

Question

I would like to know if the following is true. If $$\lim_{z\to \infty} 1/f(z) = \infty$$ is that equivalent to $$\lim_{z\to \infty} f(z) = 0?$$

Answer 1

$ \lim_{z\to\infty}1/f(z)=\infty $ means that, for any $M>0$, there exists $N\in\mathbb N$ so that, when $z>N$, $1/f(z)>M$. Therefore $f(z)<1/M$ for all $z>N$. That means that, as $z\to\infty$, we can always shrink $f(z)$ as small as possible, so yes, $\lim_{z\to\infty}f(z)=0$.

I'm assuming here that $z\in\mathbb R$. Something needs tweaking otherwise.

EDIT: the question has been changed, so yes, as @T. Bongers says below, the converse does not hold.

Answer 2

The former does imply the latter: inversion is continuous everywhere except $0$. Thus, the limit of the inverse is the inverse of the limit:

$$ \lim f(x) = \frac{1}{\lim 1/f(x)} = \frac{1}{+\infty} = 0 $$

Inversion is not defined at $0$, however, so we cannot hope to argue in the other direction:

$$ \lim \frac{1}{f(x)} ?= \frac{1}{\lim f(x)} ?= \frac{1}{0} = ??$$

However, it is true that if $\lim f(x) = 0$ then the limit points of $1/f(x)$ are $-\infty$, $+\infty$ or both. So if $\lim 1/f(x)$ exists, then it is either $+\infty$ or $-\infty$. Similarly, $\lim 1/f(x)^2$ always exists, and equals $+\infty$.