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This question already has an answer here:

How to find the limit of: $$ \lim_{x\to \infty }x\left(\sqrt[x]{a}-1\right)$$ Without using L'hôspital rule.

I've tried to bound the term and use the squeeze theorem but I couldn't find the right upper bound. I've also tried to convert $a^\frac{1}{x}$ to $e^{\frac{1}{x}\ln{a}}$ but it didn't help me.

Whats the right way to evaluate that limit?

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marked as duplicate by egreg, M Turgeon, Claude Leibovici, user63181, Yiorgos S. Smyrlis Mar 27 '14 at 13:53

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  • $\begingroup$ $\ln a$. Taylor expand. $\endgroup$ – Chris Gerig Jan 13 '14 at 9:22
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$$ \lim_{x\to +\infty } x \left(a^{1/x}-1\right) = \lim_{x\to +\infty } (\ln a) \frac{e^{(\ln a)/x} - 1}{(\ln a)/x} = (\ln a)\lim_{u \to 0} \frac{e^u - 1}{u} = (\ln a)f'(0) = \ln a,$$ where $f(u) = e^u$, and we've made the substitution $u = (\ln a)/x$ in the second limit.

Edit: Alternatively, and perhaps more simply,

$$ \lim_{x\to +\infty } x \left(a^{1/x}-1\right) = \lim_{x\to +\infty } \frac{a^{1/x} - 1}{1/x} = \lim_{u \to 0} \frac{a^u - 1}{u} = g'(0) = \ln a,$$ where $g(u) = a^u$.

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  • $\begingroup$ Perhaps in your last line you mean $g(u)$? $\endgroup$ – preferred_anon Jan 13 '14 at 10:04
  • $\begingroup$ Yes, I'll edit my answer to correct that. Thanks. $\endgroup$ – user120974 Jan 13 '14 at 10:05
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Setting $\displaystyle \frac1x=h$

$$\lim_{n\to\infty}x(\sqrt[x]a-1)=\lim_{h\to0}\frac{a^h-1}h=\ln a\lim_{h\to0}\frac{e^{h\ln a}-1}{h\ln a}=\cdots$$

Use Proof of $ f(x) = (e^x-1)/x = 1 \text{ as } x\to 0$ using epsilon-delta definition of a limit

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This limit is called Halley’s formula. For each fixed $a$, there exists an $N$ such that $x > N$ implies $\left|\frac{\log(a)}{x}\right|<1$. Note that if $|a|<1$ $$\begin{align}e^a&=1+a+\frac{1}{2}a^2+...\\&\leq1+a+a^2(\frac{1}{2}+\frac{1}{3!}|a|+\frac{1}{4!}|a|^2+...)\\&\leq1+a+a^2(1+|a|+|a|^2+...)\\&=1+a+\frac{a^2}{1-|a|}.\end{align}$$ On the other hand, modifying the above estimates, we have $$\begin{align}e^a&=1+a+\frac{1}{2}a^2+...\\&\geq1+a-a^2(\frac{1}{2}+\frac{1}{3!}|a|+\frac{1}{4!}|a|^2+...)\\&\geq1+a-a^2(1+|a|+|a|^2+...)\\&=1+a-\frac{a^2}{1-|a|}.\end{align}$$ Now write $e^{\frac{\log(a)}{x}}$ as a series in $\frac{\log(a)}{x}$ to obtain: $$1+\frac{\log(a)}{x}-\frac{\left(\frac{\log(a)}{x}\right)^2}{1-\left|\frac{\log(a)}{x}\right|}<e^{\frac{\log(a)}{x}}<1+\frac{\log(a)}{x}+\frac{\left(\frac{\log(a)}{x}\right)^2}{1-\left|\frac{\log(a)}{x}\right|}$$ and so $$-\frac{\frac{\log^2(a)}{x}}{1-\left|\frac{\log(a)}{x}\right|}<x(e^{\frac{\log(a)}{x}}-1)-\log(a)<\frac{\frac{\log^2(a)}{x}}{1-\left|\frac{\log(a)}{x}\right|}$$ Thus the result follows from the squeeze theorem, $\frac{\log^2(a)}{x}\to0$ as $x\to\infty$.

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