calculus - The limit as $n$ approaches infinity of $n(a^{1/n}-1)$

I need to know how to calculate this without using l'hospitals rule:

limit as $x$ approaches infinity of: $$x(a^{1/x}-1)$$

I saw that the answer is $\log(a)$, but I want to know how they got it.

The book implies that I should be able to find it by just using algebraic manipulation and substitution.

Answer

METHOD 1:

$$\begin{align}
\lim_{x\to \infty}x(a^{1/x}-1)&=\lim_{x\to 0^{+}}\frac{a^{x}-1}{x}\tag 1\\\\
&=\lim_{x\to 0^{+}}\frac{e^{x\log a}-1}{x}\\\\
&=\lim_{x\to 0^{+}}\frac{\left(1+(\log a)x+O( x^2)\right)-1}{x}\\\\
&=\lim_{x\to 0^{+}}\left(\log a+O(x)\right)\\\\
&=\log a

\end{align}$$

---

METHOD 2:

Another way to do this is to substitute $y=a^x$ in $(1)$. Then

$$\begin{align}
\lim_{x\to \infty}x(a^{1/x}-1)&=\lim_{x\to 0^{+}}\frac{a^{x}-1}{x}\\\\

&=\lim_{y\to 1^{+}}\frac{y-1}{\log y/\log a}\\\\
&=\log a\,\lim_{y\to 1^{+}}\frac{y-1}{\log y}
\end{align}$$

Noting that for $y>1$, $\frac{y-1}{y}\le\log y\le y-1$. Then,

$$1\le\frac{y-1}{\log y}\le y$$

and the squeeze theorem does the rest!