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

Solution 1:

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!

Solution 2:

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.