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$$\lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$$
This is one of the problems i have to solve so that i could join college. I tried using integration by parts, i tried using notations but nothing works. If someone could please help me i would deeply appreciate it. ! thanks in advance ! I know the answer to the limit is 1. But i need help proving it.
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 ${\displaystyle\int}_0^1 n ( \cos(x) - \sin(x) )^n\ dx \to 1$.
