The Limit of $x\left(\sqrt[x]{a}-1\right)$ as $x\to\infty$. 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?
 A: $$ \lim_{x\to +\infty } x \left(a^{1/x}-1\right) = \lim_{x\to +\infty } (\ln a) \frac{e^{(\ln a)/x} - 1}{(\ln a)/x} = (\ln a)\lim_{u \to 0} \frac{e^u - 1}{u} = (\ln a)f'(0) = \ln a,$$
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 $g(u) = a^u$.
A: 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 $x\to\infty$.
A: 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
