349. Another interesting limits.

by Virgil Nicula, Jul 18, 2012, 5:56 PM

PP1. Ascertain $\lim_{n\to\infty}\frac {\ln \mathrm C_{2n}^n}{2n}$ .

Proof. $C_{2n}^n=\frac {(n+1)\cdot (n+2)\cdot (n+3)\cdot\ \ldots\ \cdot (2n-1)\cdot (2n)}{1\cdot 2\cdot 3\ \cdot\ \ldots\ (n-1)\cdot n}=$ $\prod_{k=1}^n\frac {n+k}{k}\implies$

$\ln C_{2n}^n =\ln \prod_{k=1}^n\frac {n+k}{k}=$ $\sum_{k=1}^n\ln\frac {1+\frac kn}{\frac kn}\implies$ $\lim_{n\to\infty}\frac {\ln \mathrm C_{2n}^n}{2n}=$ $\lim_{n\to\infty}\frac {1}{2n}\cdot\sum_{k=1}^n\ln\frac {1+\frac kn}{\frac kn}=$ $\frac 12\cdot\int_0^1\ln\frac {x+1}{x}\ \mathrm {dx}=$

$\frac 12\cdot \left(x\ln\left|\frac {x+1}{x}\right|_0^1+\int_0^1\frac {1}{x+1}\ \mathrm{dx}\right)=$ $\frac 12\left(\ln 2+\ln 2\right)\implies$ $\boxed{\lim_{n\to\infty}\frac {\ln \mathrm C_{2n}^n}{2n}=\ln 2}$ . I used $\lim_{x\searrow 0}x\ln\frac {x+1}{x}=$

$\lim_{x\searrow 0}x\ln x=0$ and the following partial integration : $\left\{\begin{array}{ccc} u(x)=\ln\frac {x+1}{x} & \implies & u'(x)=-\frac {1}{x(x+1)}\\\\ v'(x)=1 & \implies & v(x)=x\end{array}\right\|$ .


PP2. Evaluate the limit $\lim_{n\to\infty}\frac 1n\cdot\ln\frac {n!}{n^n}$ .

Proof. $\lim_{n\to\infty}\frac 1n\cdot\ln\frac {n!}{n^n}=$ $\lim_{n\to\infty}\frac 1n\cdot\sum_{k=1}^n\ln\frac kn=$ $\int_0^1\ln x\ \mathrm{dx}=$ $\left|\left(x\ln x-x\right)\right|_0^1=-1\ .$ I used the limit $\lim_{x\searrow 0}x\ln x=0$ .


PP3. Evaluate the limit $\lim_{x_ \to \frac{\pi}{3}}\frac{\sin \left(x-\frac{\pi}{3}\right)}{4\cos^2x-1}$ .

Proof. $\lim_{x_ \to \frac{\pi}{3}}\frac{\sin \left(x-\frac{\pi}{3}\right)}{4\cos^2x-1}=$ $\lim_{x_ \to \frac{\pi}{3}}\frac{\sin \left(x-\frac{\pi}{3}\right)}{(2\cos x-1)(2\cos x+1)}\stackrel{\left(t=x-\frac{\pi}{3}\right)}{\ =\ }$ $\frac 12\cdot \lim_{t\to 0}\frac{\sin t}{2\cos\left(t+\frac {\pi}{3}\right)-1}=$ $\frac 12\cdot \lim_{t\to 0}\frac{\sin t}{\cos t-\sqrt 3\sin t-1}=$ $-\frac 12\cdot \lim_{t\to 0}\frac{\frac {\sin t}{t}}{\sqrt 3\cdot \frac {\sin t}{t}+\frac {1-\cos t}{t}}\implies$

$\boxed{\lim_{x_ \to \frac{\pi}{3}}\frac{\sin \left(x-\frac{\pi}{3}\right)}{4\cos^2x-1}=-\frac {\sqrt 3}{6}}$ . Remark. I"ll use the remarkable limit $\lim_{x\to 0}\frac {\sin x}{x}=1$ and $\lim_{x\to 0}\frac {1-\cos x}{x^2}=\frac 12\implies \lim_{x\to 0}\frac {1-\cos x}{x}=0$ .
This post has been edited 9 times. Last edited by Virgil Nicula, Nov 17, 2015, 1:14 PM

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