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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 

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