<s>[INST] 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 ? 
 [/INST] 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
$$2\cdot n^3 \ge (n+1)^3=n^3+3n^2+3n+1\iff n^3 \ge 3n^2+3n+1 \quad \square$$
</s>