Infinite Limit Problems Can some one help me to solve the problems?


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*$$\lim_{n\to\infty}\sum_{k=1}^n\left|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}\right|$$

*$$ \lim_{x\to\infty}\left(\frac{3x-1}{3x+1}\right)^{4x}$$

*$$\lim_{n\to\infty}\left(1-\frac1{n^2}\right)^n $$
(Original scan of problems at http://i.stack.imgur.com/4t12K.jpg)
This questions are from ISI kolkata Computer science Phd exam entrance.The link of the original questions http://www.isical.ac.in/~deanweb/sample/MMA2013.pdf

Some slightly different approaches: For the first one
$$e^{2\pi ik/n}-e^{2\pi i(k-1)/n}=\cos\frac{2\pi k}n+i\sin\frac{2\pi k}n-\cos\frac{2\pi(k-1)}n-i\sin\frac{2\pi(k-1)}n=$$
$$=-2\sin\frac{(2k-1)\pi}n\sin\frac{\pi}n-2i\sin\frac\pi n\cos\frac{(2k-1)\pi}n\implies$$
$$\left|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}\right|=\sqrt{4\sin^2\frac{(2k-1)\pi}n\sin^2\frac{\pi}n+4\sin^2\frac\pi n\cos^2\frac{(2k-1)\pi}n}=$$
$$=2\sin\frac\pi n\implies$$
$$\sum_{k=1}^n\left|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}\right|=2n\sin\frac\pi n=2\pi\frac{\sin\frac\pi n}{\frac\pi n}\xrightarrow[n\to\infty]{}2\pi\cdot 1=2\pi$$
$$\color{red}{-------------o--------------}$$
For the second one we have
$$\left(\frac{3x-1}{3x+1}\right)^{4x}=\left[\left(1-\frac2{3x+1}\right)^{3x+1}\right]^{4/3}\cdot\left(1-\frac2{3x+1}\right)^{-4/3}\xrightarrow[x\to\infty]{}\left(e^{-2}\right)^{4/3}\cdot 1=e^{-8/3}$$
$$\color{red}{-------------o--------------}$$
For the last one we use the straightforward (try to produce the easy proof)
Lemma: If $\,\{a_n\}_{n\in \Bbb N}\;$ is a positive real sequence (enough to require "positive except perhaps a finite number of elements) and
$$\lim_{n\to\infty}a_n=L>0\;,\;\;\text{then}\;\;\lim_{n\to\infty}\sqrt[n]{a_n}=1$$
Thus we get
$$\left(1-\frac1{n^2}\right)^n=\sqrt[n]{\left(1-\frac1{n^2}\right)^{n^2}}\xrightarrow[n\to\infty]{}1$$
because
$$\left(1-\frac1{n^2}\right)^{n^2}\xrightarrow[n\to\infty]{}e$$
The above last problem, of course, is just what André wrote.
