Find the Limit $\lim_{n \rightarrow \infty}\frac{1}{(n+1) \log (1+\frac{1}{n})}$ Find the limit
$$\lim\limits_{n \rightarrow \infty}\frac{1}{(n+1) \log (1+\frac{1}{n})}$$
 A: $$\lim\limits_{n \rightarrow \infty}\frac{1}{(n+1) \log (1+\frac{1}{n})}=\lim\limits_{n \rightarrow \infty}\frac{1}{\log\bigg[ (1+\frac{1}{n})^{n+1}\bigg]}$$
Now, use 
$$\lim_n  (1+\frac{1}{n})^{n+1} =e$$
A: Notice, $$\lim_{n\to \infty}\frac{1}{(n+1)\log\left(1+\frac{1}{n}\right)}$$ $$=\lim_{n\to \infty}\frac{1}{n\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{n}\right)}$$ $$=\lim_{n\to \infty}\frac{\left(\frac{1}{n}\right)}{\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{n}\right)}$$  Let $\frac{1}{n}=t \implies t\to 0\ as \ n\to \infty$ $$=\lim_{t\to 0}\frac{t}{\left(1+t\right)\log\left(1+t\right)}$$ Now, applying L-hospital's rule for $\frac{0}{0}$ form  $$=\lim_{t\to 0}\frac{\frac{d(t)}{dt}}{\frac{d}{dt}\left(\left(1+t\right)\log\left(1+t\right)\right)}$$
$$=\lim_{t\to 0}\frac{1}{\left(1+t\right)\frac{1}{(1+t)}+(1)\log\left(1+t\right)}$$ $$=\lim_{t\to 0}\frac{1}{1+\log(1+t)}$$ $$=\frac{1}{1+\log(1+0)}$$ $$=\frac{1}{1+0}$$ $$=\frac{1}{1}=\color{blue}{1}$$
A: $$\lim_{n\to\infty} \frac1{(n+1)\ln(1+\frac1n)} =\lim_{n\to\infty} \frac{(n+1)^{-1}}{\ln(1+\frac1n)}$$
Using l'Hopital's rule $\big(\frac00\big)$:
$$\lim_{n\to\infty}\frac{(n+1)^{-2}}{\frac{1}{n^2}\big(\frac{n}{n+1}\big)}
= \lim_{n\to\infty}\frac{n}{n+1} = 1$$
A: Set $1/n=h\implies h\to0^+$ 
$$\lim\limits_{n \rightarrow \infty}\frac{1}{(n+1) \log\left(1+\frac1n\right)}$$
$$=\lim_{h\to0^+}\dfrac h{(h+1)\ln(1+h)}$$
$$=\dfrac1{\lim_{h\to0^+}\dfrac{\ln(1+h)}h}\cdot\dfrac1{\lim_{h\to0^+}(1+h)}=?$$
A: HINT: $$\frac{1}{(n+1)\log\left(1+1/n\right)}=\frac{1}{\log\left(1+1/n)^n(1+1/n)\right)}=\frac{1}{\log(1+1/n)^n+\log(1+1/n)}$$
