limit ( n* ( a^(1/n)-1), n tends to infinity)
Thx, mate!
Substitute $\displaystyle u=\frac1n,$
$\displaystyle \mathop {\lim }\limits_{n \to \infty } n\left( {a^{1/n} - 1} \right) = \mathop {\lim }\limits_{u \to 0} \frac{{a^u - 1}}
{u},$
and this is fairly well known, the answer is $\displaystyle \ln a$ for $\displaystyle a>0.$ Substitute $\displaystyle v=\frac1{a^u-1}.$
Just in case you were wondering why it is natural log, let me show you using L'Hopital's Rule.
When the limit ends up being in the form of $\displaystyle \frac{\infty}{\infty}$ OR $\displaystyle \frac{0}{0}$, then take the derivative of the top over the derivative of the bottom: $\displaystyle \frac{f'(x)}{g'(x)}$ and take the limit...
$\displaystyle lim \frac{a^u-1}{u} = lim \frac{a^u ln|a|}{1} \Rightarrow ln|a|$