Showing that $ \lim_{x \to \infty} x(1/2)^x = 0$ Can someone explain to me why $$ \lim\limits_{x \to \infty} x\bigg(\frac{1}{2}\bigg)^x = 0$$ 
Is it because the $\big(\frac{1}{2} \big)^x$ goes towards zero as $ x $ approaches $\infty$, and anything multiplied by $0 $ included $\infty$ is $0$ ?
Or does this kind of question require using l'hopital's rule because it is in the form (0*$\infty$)?
I thought it could be solved this way. Please let me know if it is correct:
$$ \lim\limits_{x \to \infty} \bigg(\frac{x}{\frac{1}{2^{-x}}}\bigg)$$
L'hopitalize the above and get:
$$ \lim\limits_{x \to \infty} \bigg(\frac{1}{x(2^{x-1})}\bigg) = 0$$

You'd better write the process as $x \to +\infty$. Thus$$\lim_{x \to +\infty} x \left(\frac{1}{2}\right)^x=\lim_{x \to +\infty} \frac{x}{2^x}=\lim_{x \to +\infty} \frac{1}{\ln 2\cdot2^x}=\frac{1}{+\infty}=0. $$
