Question

Limita cand n tinde la infinit din n*rad(n+1)*( e^1/rad(n)-e^1/rad(n+1) )

Answer 1

$\lim\limits_{n\to \infty}\left[n\cdot\sqrt{n+1}\cdot\left(e^{\frac{1}{\sqrt{n}}}-e^{\frac{1}{\sqrt{n+1}}}\right)\right] =$

$= \lim\limits_{n\to \infty}\left[n\cdot\sqrt{n+1}\cdot e^{\frac{1}{\sqrt{n+1}}}\left(e^{\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}}-1\right)\right]$

$= \lim\limits_{n\to \infty}\left[n\cdot\sqrt{n+1}\cdot e^{\frac{1}{\sqrt{n+1}}}\cdot \dfrac{e^{\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}}-1}{\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}}\cdot \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\right]$

$= \lim\limits_{n\to \infty}\left[n\cdot\sqrt{n+1}\cdot 1\cdot 1\cdot \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\right]$

$= \lim\limits_{n\to \infty}\dfrac{n\cdot\sqrt{n+1}\cdot \left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n}\cdot \sqrt{n+1}}$

$= \lim\limits_{n\to \infty}\left[\sqrt n \cdot \left(\sqrt{n+1}-\sqrt{n}\right)\right]$

$= \lim\limits_{n\to \infty}\left(\sqrt n \cdot \dfrac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\right)$

$= \lim\limits_{n\to \infty} \dfrac{\sqrt n}{\sqrt{n+1}+\sqrt{n}}$

$= \lim\limits_{n\to \infty} \dfrac{\sqrt n}{\sqrt{n}\cdot \left(\dfrac{\sqrt{n+1}}{\sqrt{n}}+1\right)}$

$= \lim\limits_{n\to \infty} \dfrac{1}{\sqrt{\dfrac{n+1}{n}}+1}$

$= \dfrac{1}{1+1}$

$= \boxed{\dfrac{1}{2}}$