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This question already has an answer here:

How to find limit of

$$\lim_{n\rightarrow \infty} n(x^{1/n}-1)$$

Via L'Hopital's rule

$$\lim_{n\rightarrow \infty} n(x^{1/n}-1) = \lim_{n\rightarrow \infty} \frac{1}{n} x^{\frac{1}{n}-1} / \frac{-1}{n^2} =\lim_{n\rightarrow \infty} x^{\frac{1}{n}-1} / \frac{-1}{n}$$

but that doesn't help.

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marked as duplicate by user91500, Michael Albanese, Claude Leibovici, Kirill, egreg Jul 7 '14 at 10:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Put $y_n=x^{1/n}-1$ and $1/n=log_x(y_n+1)$ $\endgroup$ – pointer Jul 7 '14 at 8:06
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Your application of L'Hopital rule is incorrect. You have to differentiate with respect to $n$ instead of $x$ which you have done in the numerator.

$$\lim_{n\rightarrow \infty} n(x^{1/n}-1)=\lim_{h\rightarrow 0} \frac{x^h-1}{h}=\lim_{h\rightarrow 0} x^h\ln x=\ln x$$

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Rewrite the limit to $$\lim\limits_{h\rightarrow 0^+}{\frac{x^h-1}{h}}$$ Why don't you now use L'Hôpital's rule?


If you know that $\lim\limits_{h\rightarrow 0}{\dfrac{e^h-1}{h}}=1$, instead of L'Hôpital's rule, you can think about the substiution $x^h=e^{h\ln{x}}$.

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  • $\begingroup$ and remember that $x^h=e^{h\log(x)}$ $\endgroup$ – Claude Leibovici Jul 7 '14 at 8:00
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Use the derivative definition of $x^t$ (if you are allowed to, it is $x^t \log x$): $$ \lim_{n \to \infty} n (x^{\frac{1}{n}} -1) = \lim_{n \to \infty} \frac{x^{\frac{1}{n}}-1}{\frac{1}{n}} = \lim_{h \to 0}\frac{x^h -1}{h} = \lim_{h \to 0}\lim_{t \to 0}\frac{x^{t+h} - x^t}{h} = \lim_{t \to 0}\lim_{h \to 0} \frac{x^{t+h} - x^t}{h}\\ =\lim_{t \to 0} x^t \log x = \log x $$

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