Yes it does. And we needn't know about $\lim_{x\to\infty} f(x)$ as it is not important for the behavior of $f$ near $0$ (and as $x \mapsto \frac 1x$ maps $[n,\infty)$ to $(0, \frac 1n]$ the second limits also talks about the behavior of $f$ near 0 [and that's the proof]).
To be more exact, let $\epsilon > 0$. Then by existence of $a :=\lim_{x\to 0^+} f(x)$ there is an $\delta > 0$ such that $|f(x) - a| < \epsilon$ for $0 < x < \delta$. Let $M := \frac 1\delta$, for $x > M$ we have $0 < \frac 1x < \frac 1\delta$, hence \[ \left|f\left(\frac 1x\right) - a \right| < \epsilon \] This proves $\lim_{x\to\infty} f(\frac 1x) = a = \lim_{x\to 0^+} f(x)$.