Evaluating $\lim\limits_{x\to \infty}\frac{x^2}{2^x-1}$

Question

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 ?

Answer 1

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

$$2\cdot n^3 \ge (n+1)^3=n^3+3n^2+3n+1\iff n^3 \ge 3n^2+3n+1 \quad \square$$

Answer 2

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.

Answer 3

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.