Evaluating $\lim\limits_{x\to \infty}\frac{x^2}{2^x-1}$ 
I would like to evaluate
  $$\lim_{x\to \infty}\frac{x^2}{2^x-1}$$
  Without using L'HOSPITAL's rule nor series. 

I tried more than one 
technique such the sub 
$$x=\frac{1}{y}$$ 
But i could not get the solution ? 
 A: Using the hint provided by Jyrki Lahtonen in the comments, we may sneakily apply the binomial theorem in order to obtain
$$ 2^n - 1 = (1+1)^n - 1 = -1 + \sum_{k=0}^{n} \binom{n}{k} \ge \binom{n}{k} $$
for any choice of natural number $k$ less than $n$.  This implies that
$$ \frac{n^2}{2^n - 1} \le \frac{n^2}{\binom{n}{k}}. \tag{1}$$
By choosing $n$ large enough, it is possible to find a value of $k$ such that $\binom{n}{k} \gg n^2$.  For example, if $k = 3 < n$, then we have
$$ \binom{n}{3} = \frac{n!}{(n-3)!3!} = \frac{n(n-1)(n-2)}{3\cdot 2}. $$
Plugging this into (1), we get
$$\frac{n^2}{2^n - 1}
\le \frac{n^2}{\binom{n}{3}}
= 6 \frac{n^2}{n(n-1)(n-2)}
= \frac{6n^2}{n^3 - 3n^2 + 2n}.
$$
The expression that we want to limit is positive for any $n > 0$, thus applying the squeeze theorem we can take limits to get
$$ 0 \le
\lim_{n\to\infty} \frac{n^2}{2^n - 1}
\le \lim_{n\to\infty} \frac{6n^2}{n^3 - 3n^2 + 2n}
= \lim_{n\to \infty} \frac{\frac{6}{n}}{1 - {\frac{3}{n}} + \frac{2}{n^2}}
= 0.
$$
(The last expression, obtained via multiplication by $\frac{1/n^3}{1/n^3}$ is perhaps not strictly necessary, but I think that it aids in understanding—more generally, we might simply recall that the limit (at infinity) of a rational expression is 0 if the degree of the numerator is smaller than the degree of the denominator; $\infty$ if the numerator has greater degree, and some nonzero constant if the numerator and denominator have the same degree.)
EDIT:  I left off a step of the computation that seemed obvious to me, but for the sake of completeness, all that has been shown above is that if
$$ \lim_{x\to \infty} \frac{x^2}{2^x - 1} $$
exists, then it must be equal to 0.  To finish the argument, we might note that
$$ \frac{x^2}{2^x - 1}
\le \frac{\lceil x \rceil^2}{2^{\lfloor x \rfloor} - 1}
= \frac{(\lfloor x \rfloor + 1)^2}{2^{\lfloor x \rfloor} - 1}
= \frac{\lfloor x \rfloor^2}{2^{\lfloor x \rfloor} - 1} + \frac{2\lfloor x \rfloor + 1}{2^{\lfloor x \rfloor} - 1}. $$
The first term is exactly the one analyzed with $n = \lfloor x \rfloor$.  The second term can by analyzed similarly.
Alternatively, with
$$ f(x) = \frac{x^2}{2^x - 1}, $$
look at $f'(x)$, and note that $f'(x) < 0$ for $x$ sufficiently large (using the same kinds of estimates as above; this requires a little work, but is not unreasonably difficult).  Since the derivative is negative, the function is decreasing.  Moreover, the function is nonnegative for all $x > 1$, and so bounded below by zero.  As $f(x)$ is monotonically decreasing and bounded, it must have a limit.  By the above argument, the limit is zero.
A: Enforcing the changes of variables $ u= \frac{x\ln(2)}{2}$ gives $$\lim_{x\to \infty}\frac{x^2}{2^x-1} =\lim_{x\to \infty}\frac{x^2}{2^x}\cdot\frac{1}{1-\frac{1}{2^x}}= \lim_{x\to \infty}x e^{-x\ln(2)}\\= \lim_{x\to \infty}\frac{4}{\ln^2 2}\left(\frac{x\ln(2)}{2}e^{-\frac{x\ln(2)}{2}}\right)^2 =\lim_{u\to \infty} \frac{4}{\ln^2 2}\left(ue^{-u}\right)^2  =0$$
se also here  https://math.stackexchange.com/q/2587440.
A: Note that
$$\frac{x^2}{2^x-1}=\frac{2^x}{2^x-1}\cdot \frac{x^2}{2^x}\to1\cdot0=0$$
indeed
$$\frac{2^x}{2^x-1}=\frac{1}{1-\frac{1}{2^x}}\to 1$$
and since eventually, notably for $x>10$, we have that $2^x>x^3$
$$\frac{x^2}{2^x}<\frac{x^2}{x^3}=\frac1x\to0$$
EDIT
Proof by induction of $2^n>n^3$.
Base case
$n=10\implies2^{10}=1024>10^3=1000$
Induction step
We need to prove that
$2^n>n^3\implies 2^{n+1}>(n+1)^3$
Let observe that
$$2^{n+1}=2\cdot 2^n\stackrel{\text{inductive hypothesis}}\ge 2\cdot n^3 \stackrel{\text{?}}\ge (n+1)^3$$
and the last inequality is true since
$$2\cdot n^3 \ge (n+1)^3=n^3+3n^2+3n+1\iff n^3 \ge 3n^2+3n+1 \quad \square$$
