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How can I find such a limit: $$\lim_{n\to\infty}n(x^{1/n}-1)$$ I tried using a kind of binomial formulas. But nothing helped so far.

Thank you!

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marked as duplicate by Maisam Hedyelloo, David Mitra, Danny Cheuk, Lord_Farin, Amzoti Jun 28 '13 at 22:16

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.

Write the limit as $n(x^{1/n}-1) = \frac{(x^{1/n}-1)}{1/n}$, make substitution $t= 1/n$, this is the definition of the derivative at zero for $d(x^t)/dt$ $$ \lim_{t\to 0^+}\frac{(x^{t}-x^0)}{t-0} = \frac{dx^t}{dt} $$

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Hint: We have $x^{1/n}=e^{\log x/n}$. Now use the first two terms of the Maclaurin series of $e^t$.

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Thank you guys! – Eu2718 Jun 28 '13 at 22:07

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