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Find the limit of the sequences: $a_n = \left(1-\frac{1}{n}\right)^n $, $a_n = \sqrt{n+1}-\sqrt{n}$

In the following problems, we find the limits of sequences $\{a_n\}$ given the functional form of the sequences. We assume that some fundamental limits such as $$ \lim_{n\rightarrow\infty}\frac{1}{n} = 0 $$ and $$ \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n \equiv e $$ are known.
  1. $a_n = \left(1-\frac{1}{n}\right)^n $ $$\begin{align} a_n &= \left(\frac{n-1}{n}\right)^n% = \left(\frac{n}{n-1}\right)^{-n}\\ &=\left(\frac{n-1}{n-1}+\frac{1}{n-1}\right)^{-n}\\ &=\left(1+\frac{1}{n-1}\right)^{-n}\\ &=\left(1+\frac{1}{n-1}\right)^{-(n-1)}\left(1+\frac{1}{n-1}\right)^{-1}\\ &=\frac{\left(1+\frac{1}{n-1}\right)^{-1}}{\left(1+\frac{1}{n-1}\right)^{(n-1)}}\\ % \lim_{n\rightarrow\infty}a_n% &=\lim_{n\rightarrow\infty}\frac{\left(1+\frac{1}{n-1}\right)^{-1}}{\left(1+\frac{1}{n-1}\right)^{(n-1)}}\\ &=\frac{\lim_{n\rightarrow\infty}\left(1+\frac{1}{n-1}\right)^{-1}}{\lim_{{n-1}\rightarrow\infty}\left(1+\frac{1}{n-1}\right)^{(n-1)}}\\ &=\frac{1}{e} = e^{-1}.□ \end{align}$$
  2. $a_n = \sqrt{n+1}-\sqrt{n}$ $$\begin{align} a_n% &= \left(\sqrt{n+1}-\sqrt{n}\right)\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\ & = \frac{1}{\sqrt{n+1}+\sqrt{n}}\\ \lim_{n\rightarrow\infty}a_n% &= \lim_{n\rightarrow\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} = 0.□ \end{align}$$

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