- Thread starter
- #1

- Thread starter Amer
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- Thread starter
- #1

- Feb 13, 2012

- 1,704

May be that the most comfortable solution is to write...how to solve this limit

[tex]\displaystyle\lim_{x\rightarrow \infty} \dfrac{\sqrt[x+1]{x+1}-1}{\sqrt[x]{x}-1}[/tex]

(Latex Question :the x of the root is not clear how to make it better )

$\displaystyle \sqrt[1+x]{1+x}= e^{-\frac{\ln (1+x)}{1+x}}$

$\displaystyle \sqrt[x]{x}= e^{- \frac{\ln x}{x}}$

...and then apply l'Hopital's rule...

Kind regards

$\chi$ $\sigma$

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- #3

- Apr 3, 2012

- 37

Alternative:how to solve this limit

[tex]\displaystyle\lim_{x\rightarrow \infty} \dfrac{\sqrt[x+1]{x+1}-1}{\sqrt[x]{x}-1}[/tex]

You can multiply each of the four terms by [tex]\dfrac{1}{\sqrt[x + 1]{x + 1}}[/tex]