# Evaluating $\lim_{n\rightarrow \infty}n\int_{0}^{1}\left(\cos x-\sin x\right)^n\text{ d}x$

$$\lim_{n\rightarrow \infty}n\int_{0}^{1}\left(\cos x-\sin x\right)^n\text{ d}x$$

A) $\infty$

B) $0$

C) $1$

D) $\frac{1}{2}$

E) $\cos 1$

Source: admission 2015 Technical University of Cluj-Napoca

I tried with this result, but I'm not if it holds

If $f:[a,b]\rightarrow R$ f is differentiable with derivative nonzero, and $\displaystyle \lim_{n\to \infty}\int_a^b f^n(x)dx=0$ then $\displaystyle \lim_{n\to\infty} n \cdot \int_a^b f^n(x)dx=\lim_{n\to\infty} \left(\frac{f^n(b)}{f'(b)}-\frac{f^n(a)}{f'(a)}\right)$

I used $f(x) = cosx - sinx$ but I'm not sure if $\displaystyle \lim_{n\to \infty}\int_a^b f^n(x)dx=0$

I have tried with this too, $\cos x-\sin x= \frac{2}{\sqrt{2}}\cos (x+\frac{\pi}{4})$, but at one moment I can't continue... I can find a recurrence formula, but it's very complicated and I can't see the limit...

• Any tries, thoughts or comments yourself on the problem? – mickep Nov 22 '15 at 20:12
• I have tried with this, but I'm not sure if it's correct <br> $f:[a,b]\rightarrow R$ if $\displaystyle \lim_{n\to \infty}\int_a^b f^n(x)dx=0$ then $\displaystyle \lim_{n\to\infty} n \cdot \int_a^b f^n(x)dx=\lim_{n\to\infty} \left(\frac{f^n(b)}{f'(b)}-\frac{f^n(a)}{f'(a)}\right)$ – klyn Nov 22 '15 at 20:17
• I think $\cos x-\sin x= \frac{2}{\sqrt{2}}\cos (x+\frac{\pi}{4})$ will be useful. – R.N Nov 22 '15 at 20:19
• to mickep: the source of the formula is pro-didactica.ro/forum/… – klyn Nov 22 '15 at 20:48
• @user254665 Sorry to hear about your kitty. Hope he/she recovers soon. – Mark Viola Nov 23 '15 at 21:25

Let $f(x)=\cos x-\sin x$. We note that for $x\in[0,1]$, the maximum of $f(x)$ occurs at $x=0$ for which $f(0)=1$.

And for any number $0<a < \pi/4$, $x\in [a,\pi/4]$, $0<f(x)<\cos( a) -\sin (a)<1$.

Now, we choose a positive number $\delta<\pi/2-1$. Then, we write the integral of interest as the sum

\begin{align} \int_0^1 (\cos x-\sin x)^n\,dx&=\int_{0}^{\delta} (\cos x-\sin x)^n\,dx+\int_{\delta}^{\pi/2-1} (\cos x-\sin x)^n\,dx\\\\ &+\int_{\pi/2-1}^{\pi/4}(\cos x-\sin x)^n\,dx+\int_{\pi/4}^{1}(\cos x-\sin x)^n\,dx\\\\ &=\int_{0}^{\delta} (\cos x-\sin x)^n\,dx+\int_{\delta}^{\pi/2-1} (\cos x-\sin x)^n\,dx \tag 1\\\\ &+(1+(-1)^n)\int_{\pi/2-1}^{\pi/4}(\cos x-\sin x)^n\,dx \end{align}

Now, we can see that the second and third integrals on the right-hand side of $(1)$ are positive and bounded by $(1-\pi/4)(\cos \delta-\sin \delta)^n$ and $2(1-\pi/4)(\sin (1)-\cos (1))^n$, respectively. Therefore,

$$\lim_{n\to \infty}n\int_{\delta}^1(\cos x-\sin x)^n\,dx=0$$

For the first integral on the right-hand side of $(1)$, we proceed as follows. Given any $\epsilon>0$, we now choose $0<\delta<\epsilon$. Then, using the mean-value theorem, we find that for $x\in[0,\delta]$, the function $f$ satisfies the inequalities

$$1-(1+\epsilon)x \le f(x)\le 1-x \tag 2$$

Applying $(2)$ to the first integral on the right-hand side of $(1)$ yields

$$n\int_{0}^{\delta}\left(1-(1+\epsilon)x\right)^n\,dx\le n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx\le n\int_{0}^{\delta} (1- x)^n\,dx \tag 3$$

Evaluating the integrals in $(3)$ that bound $n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx$ reveals

$$\frac{n}{n+1}\frac{\left(1-(1-(1+\epsilon)\delta)^{n+1}\right)}{1+\epsilon}\le n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx\le \frac{n}{n+1}\left(1-(1-\delta)^{n+1}\right) \tag 3$$

Using the squeeze theorem ($n\to \infty$) we see that for any given $\epsilon>0$, we can choose $0<\delta <\epsilon$ so that

$$1-\epsilon\le \frac{1}{1+\epsilon}\le \lim_{n\to \infty}n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx\le 1$$

Therefore, we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx=1}$$

• Thank you! I think your solution is correct and I understand everything, excepting inequalities (2), they look correct but I'm not sure how exactly to prove... – klyn Nov 24 '15 at 5:15
• You're welcome! My pleasure. – Mark Viola Nov 24 '15 at 5:16
• As for inequality $(2)$, start with the MVT. Then use, $\sin \delta +\cos \delta<1+\delta$. – Mark Viola Nov 24 '15 at 5:18
• @DrMV I voted your solution 14 hours ago, but unfortunately I'm new, my reputation is just 9 and the vote is not visible until I don't have reputation 15, also I think I'm not a math beginner but I'm impressed by your solution Thank you again! – klyn Nov 24 '15 at 19:18