length of the curve $y=x^n$ in the unit square

Question

Let $l_n$ be the length of the curve $y=x^n$ in $[0,1]\times[0,1]$.

Then obviously $\lim_{n\to\infty}l_n = 2$.

What about $\lim_{n\to\infty}(n(2-l_n))$ ?

The formula $l_n = \int_0^1\sqrt{1+n^2x^{2n-2}}\: dx$ contains (probably) an nonelementary integral...

Answer 1

We may use a variation of Laplace's method: let $f_n(x)=\sqrt{1+n^2(1-x)^{2n-2}}$.

$f_n(x)$ decays pretty fast on $[0,1]$, and by considering the Taylor series in a right neighbourhood of the origin we have: $$ f_n(x)\approx \max\left(1,\sqrt{1+n^2}\,e^{-nx}\right)\tag{1} $$ so that: $$ l_n \approx 1+\frac{1}{n}\sqrt{1+n^2} \tag{2} $$ gives that the limit $\lim_{n\to +\infty}n(2-l_n)$ is expected to be zero. We just have to turn our approximations into real inequalities, it looks not so difficult.