The Limit of $x\left(\sqrt[x]{a}-1\right)$ as $x\to\infty$. [duplicate]

Solution 1:

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

Solution 2:

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