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$$\lim_{n\to \infty} \sqrt[n] n = \lim_{n\to \infty} n^{\frac{1}{n}} = \lim_{n \to \infty} \{(1+(n-1))^{\frac{1}{n-1}}\}{^{(n-1)\frac{1}{n}}} = \lim_{n\to \infty} e^{\frac{n-1}{n}} = e$$ But this clearly isn't true as the actual limit is $1$. Where did I go wrong?

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$$\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$$

not

$$\lim_{u\to\infty}\left(1+u\right)^{1/u}= e$$

Let $A=\lim_{u\to\infty}\left(1+u\right)^{1/u}\implies\ln A=\lim_{u\to\infty}\dfrac{\ln(1+u)}u=\lim_{u\to\infty}\dfrac1{1+u}=0$ (Using L'Hospital)

$\implies A=e^0$

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  • $\begingroup$ Interesting, i thought it was correct both ways. Thanks! $\endgroup$ – user304538 Jan 12 '16 at 10:55
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    $\begingroup$ If it is meant to use it the other way round, at least the limt must not go to infinity but to zero. I didn't check it actually but I think that should doable in a similar way as the original one $\endgroup$ – Gottfried Helms Jan 12 '16 at 11:17
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Notice, your answer is incorrect. your mistake $\lim_{n\to \infty}\left(1+(n-1)\right)^{1/(n-1)}\ne e$ . Follow the method as $$\lim_{n\to \infty}\sqrt[n]{n}=\lim_{n\to \infty}n^{1/n}=\exp\lim_{n\to \infty}\frac 1n \ln(n)$$ Applying L-Hospital's rule for $\frac {\infty}{\infty}$ form,
$$=\exp\lim_{n\to \infty}\frac{\frac{1}{n}}{1}$$ $$=\exp\lim_{n\to \infty}\frac{1}{n}$$$$=e^0=\color{red}{1}$$

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You can do it using another way $$A=n^{\frac 1n}\implies \log(A)=\frac {\log(n)}n$$ and you know that $\log(n)$ is "slower" than $n$. So, $\log(A)\to 0$ and then $A\to 1$.

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this is a wrong answer because you do (1 + inf)^(1/inf) is e but euler defined this limit by (1+(something that go to zero))^(1/the something in the base that go to zero) = e, and that for why your answer wrong.

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    $\begingroup$ Use tex format in your posts, please. $\endgroup$ – Nosrati Sep 8 '17 at 21:22

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