Question Number 160006 by HongKing last updated on 23/Nov/21
$$\mathrm{Evaluate}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{n}}} {\overset{\boldsymbol{\mathrm{n}}+\mathrm{1}} {\int}}\:\mathrm{e}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} \:\mathrm{dx}\:=\:? \\ $$$$ \\ $$
Commented by kowalsky78 last updated on 23/Nov/21
![The answer is 1. In fact, 1≤e^(1/x) ≤e^(1/n) for every x∈[n,n+1] and every n. Integrating this inequality and taking limit we conclude that the answer is 1.](https://www.tinkutara.com/question/Q160049.png)
$${The}\:{answer}\:{is}\:\mathrm{1}.\:{In}\:{fact},\:\mathrm{1}\leqslant{e}^{\frac{\mathrm{1}}{{x}}} \leqslant{e}^{\frac{\mathrm{1}}{{n}}} \:\:{for}\:{every}\:{x}\in\left[{n},{n}+\mathrm{1}\right]\:{and}\:{every}\:{n}. \\ $$$${Integrating}\:{this}\:{inequality}\:{and}\:{taking}\:{limit}\:{we}\:{conclude}\:{that}\:{the}\:{answer}\:{is}\:\mathrm{1}. \\ $$
Commented by HongKing last updated on 25/Nov/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser} \\ $$