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If $f$ is strictly growing in $(y,\infty) \subset (0,\infty)$ does it necessarily contradict the statement: $\lim_{x\to \infty}f(x)=0$?

I know how to prove it if $f$ is positive at some $y_1 \in (y,\infty)$:

Let $\epsilon = f(y_1)$.

Since $f$ is strictly growing in $(y_1,\infty)$, there exists $N_1 > 0$ for which $x>N_1 \implies f(x)>\epsilon$

Due to the fact that $\lim_{x\to\infty}f(x)=0$ there exists $N_2$ in a way that if $x>N_2 \implies |f(x)| < \epsilon$.

Let $N = \max\{N_1,N_2\}$

for every $x>N$ both $|f(x)| <\epsilon, f(x)>\epsilon$ which is a contradiction.

However is it also true if I don't know that there's a positive value in that interval?

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    You cannot fail to know that there's a positive value because *all* values are positive – 2017-01-22
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    Couldn't it be that all values will be negative? – 2017-01-22
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    Sure, look at $f(x) = -1/x$ on $(0,\infty).$ – 2017-01-22
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    Oops, I read $(y,\infty)\to(0,\infty)$ where you actually wrote $\subset$ – 2017-01-22
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    Than I'm still stuck on my previous question here http://math.stackexchange.com/questions/2108025/lim-x-to-inftyfx-0-prove-that-f-is-strictly-growing-in-0-infty/2108035#2108035 – 2017-01-22

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