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.
 A: Notice that this is the same limit as 
$$\lim_{x \to 0^+} \frac{a^x - 1}{x - 0}$$
This is one side of the definition of the derivative of $f(x) = a^x$ evaluated at $x = 0$. As $f'(x) = \ln a . a^x$ and thus $f'(0) = \ln a$, it follows that
$$\lim_{x \to 0^+} \frac{a^x - 1}{x - 0} = \ln a$$
I am all but certain there is not an evaluation of that limit using only 'algebraic manipulation and substitution', given standard definitions of the function involved.
A: 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!
A: The exponential function is a convex function, hence $x<y$ gives:
$$ e^x \leq \frac{e^y-e^x}{y-x} \leq e^{y} \tag{1}$$
and assuming $a>1$ we have:
$$ e^{0}\leq \frac{e^{x\log a}-e^0}{x\log a-0}\leq e^{x\log a}\tag{2}$$
or:
$$ \log a \leq \frac{a^x-1}{x}\leq a^x \log a \tag{3}$$
hence the claim follows by squeezing.
A: $\newcommand{\d}{\,\mathrm{d}}$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 $a<1$).
