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Find $\lim _{n \rightarrow \infty} \frac{n}{n+1}$

$\lim _{n \rightarrow \infty} \frac{n}{n+1}=\lim \frac{\infty}{\infty+1}=\frac{\infty}{\infty}$

$\lim _{n \rightarrow \infty} \frac{n}{n+1}=\lim _{n \rightarrow \infty} \frac{n(1)}{n\left(1+\frac{1}{n}\right)}$

$=\lim _{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}=\frac{1}{1+\frac{1}{\infty}}=\frac{1}{1+0}=1$

is convergent

Find $\lim _{n \rightarrow \infty} \frac{\ln (n)}{n}$

$\lim _{n \rightarrow \infty} \frac{\ln (n)}{n}=\frac{\ln (\infty)}{\infty}=\frac{\infty}{\infty}$

But we can't apply L'hopital Rule on (n) because it is discrete

so we define $f(x)=\frac{\ln (x)}{x}$

$\lim _{x \rightarrow \infty} \frac{\ln (x)}{x}=\lim _{x \rightarrow \infty} \frac{\frac{1}{x}}{1}$

$\frac{1}{\infty}=0$

$\lim _{n \rightarrow \infty} \frac{\ln (n)}{n}=0$

Evalute $\lim _{n \rightarrow \infty} \frac{(-1)^{n}}{n}$ if it exists

$\lim _{n \rightarrow \infty}\left|\frac{(-1)^{n}}{n}\right|=\lim _{n \rightarrow \infty} \frac{1}{n}=0$

$∴ \lim _{n \rightarrow \infty}\left(\frac{(-1)^{n}}{n}\right)=0$

Show that the sequence $\left\{\frac{3}{n+5}\right\}$ is decreasing

Decresing $\rightarrow a_{n+1}<a_ n$

$\frac{3}{n+1+5}<\frac{3}{n+5}$

$\frac{3}{n+6}<\frac{3}{n+5}$ True

$\left\{\frac{3}{n+5}\right\}$ is decreasing

or $f(x)=\frac{3}{x+5} \rightarrow f^{\prime}(x)=\frac{-3}{(x+5)^{2}}<0$

$\left\{\frac{3}{n+5}\right\}$ is decreasing

Show that the sequence $a_{n}=\frac{n}{n^{2}+1}$ is decreasing

$\{a n\}$ is decreasing $\Rightarrow a_{n+1}<a_{n}$

$\frac{n+1}{(n+1 )^{2}+1}<\frac{n}{n^{2}+1}$

$(n+1)\left(n^{2}+1\right)<n\left((n+1)^{2}+1\right)$

$n^{3}+n+n^{2}+1<n\left(n^{2}+2 n+1+1\right)$

$n^{3}+n^{2}+n+1<n^{3}+2 n^{2}+2 n$

$1<n^{2}+n$

$n^{2}+n>1$ is true

$a_ {n+1}<a_{n}$

$\left\{a_{n}\right\}$ is decreasing

Find the limit $\lim _{n \rightarrow \infty} \frac{\cos ^{2} n}{2^{n}}$

we knew that $0 \le \cos ^{2} n \le 1$

$\frac{0}{2^{n}} \leq \frac{ \cos ^{2} n}{2^{n}} \leq \frac{1}{2^{n}}$

$0 \leq \frac{\cos ^{2} n}{2^{n}} \leq \frac{1}{2^{n}}$

$\lim _{n \rightarrow \infty} 0=0$

$\lim _{n \rightarrow \infty} \frac{1}{2^{n}}=\frac{1}{2^{\infty}}=\frac{1}{\infty}=0$

$\lim _{n \rightarrow \infty} \frac{\cos ^{2} n}{2^{n}}=0$ by squeeze theorem