Find $\lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$

Write the integrand as $(\cos(x)-\sin(x))^n=e^{n\log(\cos(x)-\sin(x))}$. Looking at a few plots it becomes immediatly clear the integral will be dominated by a small region around the origin with width $\epsilon\sim1/n$ as $n \rightarrow\infty$. The more formal reason for that is, that the exponent is nearly zero around the origin

$$ I_n=\int_0^1(\cos(x)-\sin(x))^n\sim \int_0^{\epsilon}e^{n\log(\cos(x)-\sin(x))} $$

Taylor expansion of the exponent yields

$$ I_n\sim\int_0^{\epsilon}e^{-x n} $$

by pushing $\epsilon$ to infinity we introduce only an exponentially small error so

$$ I_n \sim\int_0^{\infty}e^{-x n}=\frac{1}{n} $$

which yields

$$ \lim_{n \rightarrow \infty} n I_n=1 $$

The following common inequalities suffice here: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

• $1+x+\lfrac12 x^2 \le \exp(x) \le 1 + x + x^2$ for every real $x \in [0,\ln(2)]$.

• $1 - \lfrac12 x^2 \le \cos(x) \le 1 - \lfrac12 x^2 + \lfrac1{24} x^4$ for every real $x$.

• $x - \lfrac16 x^3 \le \sin(x) \le x$ for every real $x \ge 0$.

They can be proven elementarily by recursively comparing derivatives.

As $n \to \infty$:

  Let $r = n^{2/3}$.

  Given any $x \in [0,\lfrac1r]$:

    $\cos(x) - \sin(x) \le 1 - x + \lfrac16 x^3 \le \exp(-x)$.

    $\cos(x) - \sin(x) \ge 1 - x - \lfrac12 x^2 \ge \exp(-x-2x^2)$ since $x \to 0$.

  Thus $(\cos(x)-\sin(x))^n \in \exp(-x) ^ n \cdot [\exp(-\frac{2}{r^2}),1]^n$.

  Thus ${\displaystyle\int}_0^{\lfrac1r} n ( \cos(x) - \sin(x) )^n\ dx \in [\exp(-\frac{2n}{r^2}),1] \cdot {\displaystyle\int}_0^{\lfrac1r} n \exp(-nx)\ dx$.

  Also ${\displaystyle\int}_0^{\lfrac1r} n \exp(-nx)\ dx = 1 - \exp(-\lfrac{n}{r}) \to 1$.

  Given any $x \in [\lfrac1r,1]$:

    $\cos(x) - \sin(x) \le 1 - x - x^2 ( \lfrac12 - \lfrac16 x - \lfrac1{24} x^2 ) \le 1 - \lfrac1r$.

    $\cos(x) - \sin(x) \ge 1 - x - \lfrac12 x^2 \ge -\frac12$.

    Thus $| n ( \cos(x) - \sin(x) )^n | \le n ( 1 - \lfrac1r )^n \to 0$.

  Thus ${\displaystyle\int}_{\lfrac1r}^1 n ( \cos(x) - \sin(x) )^n\ dx \to 0$.

  Therefore ${\displaystyle\int}_0^1 n ( \cos(x) - \sin(x) )^n\ dx \to 1$.