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:
$$ \lim\limits_{x \to \infty} \bigg(\frac{1}{x(2^{x-1})}\bigg) = 0$$

Let
$a_x = \dfrac{x}{2^x}$.
Then
$\dfrac{a_{x+1}}{a_x}
=\dfrac{\dfrac{x+1}{2^{x+1}}}{\dfrac{x}{2^x}}
=\dfrac{1+1/x}{2}
$.
Therefore,
for $x > 3$,
$\dfrac{a_{x+1}}{a_x}
=\dfrac{1+1/x}{2}
\lt\dfrac{1+1/3}{2}
=\dfrac46
=\dfrac23
$.
Therefore,
for $x > 3$,
$\begin{array}\\
\dfrac{a_x}{a_3}
&=\prod_{y=3}^{x-1}\dfrac{a_{y+1}}{a_y}\\
&\lt\prod_{y=3}^{x-1}\dfrac23\\
&=(2/3)^{x-3}\\
\text{so}\\
a_x
&\lt a_3(2/3)^3(2/3)^x\\
&= (3/8)(8/27)(2/3)^x\\
&= (1/9)(2/3)^x\\
\end{array}
$
To show 
in an elementary way
that $c^x
\to 0$
for any $0 < c < 1$,
let
$c = 1/(1+d)$
where $d > 0$.
Note:
What follows is not original.
Then,
$d = 1/c-1$
so,
by Bernoulli's inequality,
$(1+d)^x \ge 1+xd > xd$
so
$c^x
=1/(1+d)^x
\lt 1/(xd)
=1/(x(1/c-1))
=c/(x(1-c))
\to 0$
as $x \to \infty$.
With $c = 2/3$,
$(2/3)^x
\lt (2/3)/(x(1/3))
=2/x
$
so that
$a_x \lt (1/9)(2/x)
=2/(9x)
$.
