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:
$$ \lim\limits_{x \to \infty} \bigg(\frac{1}{x(2^{x-1})}\bigg) = 0$$
 A: No, you can't know the limit is $0$ simply because one factor goes to $0$ and the other goes to $+\infty$. Some limits of this form are infinity or nonzero finite numbers instead:
$$ \lim_{x\to\infty} x^2\cdot \frac1x = \infty \qquad\qquad
   \lim_{x\to\infty} x \cdot \frac1{2x} = \frac12 $$
We express the fact that taking limits factor for factor will not help us here, by saying that $0\cdot\infty$ is an indeterminate form. This really ought to be written as something like $(\to 0)\cdot(\to\infty)$, such that one wouldn't get the false impression that "indeterminate" is a property of "multiplying $0$ by infinity" (which is nonsense), but there's probably no hope of changing the traditional shorthand by now.
When the limit has an indeterminate form, you need to use other techniques than taking limits separately. L'Hospital is often useful for this -- and indeed gives the correct result here. (However, beware that you have used a wrong rule for the derivative of $1/2^{-x} = 2^x$. If you use the right rule, you get $(\log 2)2^x$, which still goes to $+\infty$, so the original limit is indeed $0$). 
In other cases, however, you need to apply clever tricks that are specific to the particular expression you're looking at. There's no one-size-fits-all approach.
A: exponential growth is faster than linear growth. hence 0
A: 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)
$.
A: You'd better write the process as $x \to +\infty$. Thus$$\lim_{x \to +\infty} x \left(\frac{1}{2}\right)^x=\lim_{x \to +\infty} \frac{x}{2^x}=\lim_{x \to +\infty} \frac{1}{\ln 2\cdot2^x}=\frac{1}{+\infty}=0. $$
A: May be, you could consider
$$y=x \left(\frac 12 \right)^x\implies \log(y)=\log(x)-x\log(2)$$ Remember that $\log(x)$ is smaller than $x$. So, when $x$ is large $$\log(y)\sim -x \log(2)\to -\infty\implies y=e^{\log(y)}\to 0$$
