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Question Number 157216 by Khalmohmmad last updated on 21/Oct/21

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}=?$$$$\underset{x\to 0}{\lim }\frac{\sqrt{\tan 4x}}{\tan \sqrt{4x}}=?$$

Commented by cortano last updated on 21/Oct/21

$$\left(2\right)\underset{x\to 0^{+}}{\lim }\frac{\sqrt{\tan 4x}}{\tan \sqrt{4x}}=\underset{x\to 0^{+}}{\lim }\frac{\frac{\sqrt{\tan 4x}}{\sqrt{4x}}}{\frac{\tan \sqrt{4x}}{\sqrt{4x}}}$$$$=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{\frac{\tan 4x}{4x}}}{\frac{\tan \sqrt{4x}}{\sqrt{4x}}}=1$$

Answered by puissant last updated on 21/Oct/21

$$1)$$$$\mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}};ln\left(\mathcal{L}\right)=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}ln\left(\frac{n!}{n^n}\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\{ln(n!)-nln\left(n\right)\}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^n}\{ln\left(k\right)-ln\left(n\right)\}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^n}ln\left(\frac{k}{n}\right)={\int }_0^{1}ln\left(x\right)dx$$$$IBP\to \{\begin{array}{l}u=lnx\\ v^{\prime }=1\end{array}\Rightarrow \{\begin{array}{l}{u}^{\prime }=\frac{1}{x}\\ v=x\end{array}$$$$\Rightarrow ln\left(\mathcal{L}\right)={\left[xlnx\right]}_0^1-{\int }_0^{1}1dx$$$$\Rightarrow ln\left(\mathcal{L}\right)=-1\to \mathcal{L}=e^{-1}=\frac{1}{e}..$$$$\therefore \because \mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}=\frac{1}{e}..$$$$$$$$...........\mathcal{L}epuissant............$$

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