The limit as $n$ approaches infinity of $n\left(a^{1/n}-1\right)$ I need to know how to calculate this without using l'hospitals rule:
limit as $x$ approaches infinity of: $$x\left(a^{1/x}-1\right)$$
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.

$\newcommand{\d}{\,\mathrm{d}}$We know: $$\log a=\int_1^a\frac{1}{x}\d x$$And the RHS is / can be defined to be a Riemann integral, which is a limit of Riemann sums over tagged partitions as the mesh of these partitions tends to zero. In particular, since the integrand is integrable, we can use any partition. Let $a>1$ - the proof is the same for $a<1$, essentially.
For every $N\in\Bbb N$ partition $[1;a]$ into $1=t_0<t_1<\cdots<t_N=a$ through $t_i:=a^{i/N}$ (similarly to this) for $0\le i\le N$. "Tag" this partition by $\xi_i:=t_{i-1}$ for $1\le i\le N$. Since the "mesh" $\max_{1\le i\le N}(t_i-t_{i-1})=a(1-a^{-1/N})\to0,\,N\to\infty$, we can say: $$\begin{align}\log a&=\lim_{N\to\infty}\sum_{i=1}^N(t_i-t_{i-1})\cdot\frac{1}{\xi_i}\\&=\lim_{N\to\infty}\sum_{i=1}^N\frac{a^{(i-1)/N}(a^{1/N}-1)}{a^{(i-1)/N}}\\&=\lim_{N\to\infty}\sum_{i=1}^N(a^{1/N}-1)\\&=\lim_{N\to\infty}N(a^{1/N}-1)\end{align}$$
Which shows the result in a nice way (I think), at least if you only care about the limit over integers. You could extend this to a result about the limit over the reals fairly straightforwardly, given the fact that $x\mapsto x(a^{1/x}-1)$ decreases in $x$ if $a>1$ (and mimic the proof in the opposite way, if $a<1$).
