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Although this question has already been asked in general ( How to find $\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1)$?) , my question is different, because I am stuck with a specific transformation step:

$$\lim\limits_{n\rightarrow \infty} n(\sqrt[n]{a}-1) $$

I've got the solution to it, but I have problems understanding this step: $$\lim\limits_{n\rightarrow \infty} n(\sqrt[n]{a}-1) = \frac{\exp\left(\frac{\log a}{n}\right)-1}{\frac{1}{n}}\\$$

I know that I can rewrite $\sqrt[n]{a}$ as $e^{(1/n)*\log(a)}= \exp\left(\frac{\log a}{n}\right)$ with $\log = \ln$.

Question: How do I get the $\frac{1}{n}$ in the denominator.

And the final step is somehow taking the $\log a$, on both the numerator and denominator.

$$ = \log a \frac{\exp\left(\frac{\log a}{n}\right)-1}{\frac{\log a}{n}}=\log a\\$$

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    $\begingroup$ $\dfrac{1}{\frac{1}{n}}=n.$ This is what you are missing here. For the final step, multiply $\log a$ in numerator and denominator (make sure $\log a \neq 0$). $\log a \cdot \dfrac{1}{\frac{\log a}{n}}=n.$ $\endgroup$ – Krish Mar 19 '15 at 12:55
  • $\begingroup$ Ah, thanks! Now it's clear! $\endgroup$ – Christoph S Mar 19 '15 at 13:49
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If you are allowed to use the derivative of $a^x$, you can immediately see the definition of $f'(0)$, just set $\frac{1}{n} = t$: $$ \lim_{t \to 0} \lim_{h \to 0} \frac{a^{t+h}-a^h}{t} = \lim_{h \to 0} \lim_{t \to 0} \frac{a^{h+t}-a^h}{t} = \lim_{h \to 0} a^h \log a = \log a $$

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