1. find this limit

limit ( n* ( a^(1/n)-1), n tends to infinity)

Thx, mate!

2. Originally Posted by szpengchao
limit ( n* ( a^(1/n)-1), n tends to infinity)
Substitute $u=\frac1n,$

$\mathop {\lim }\limits_{n \to \infty } n\left( {a^{1/n} - 1} \right) = \mathop {\lim }\limits_{u \to 0} \frac{{a^u - 1}}
{u},$

and this is fairly well known, the answer is $\ln a$ for $a>0.$ Substitute $v=\frac1{a^u-1}.$

3. Originally Posted by Krizalid
and this is fairly well known, the answer is $\ln a$ for $a>0.$ Substitute $v=\frac1{a^u-1}.$
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 $\frac{\infty}{\infty}$ OR $\frac{0}{0}$, then take the derivative of the top over the derivative of the bottom: $\frac{f'(x)}{g'(x)}$ and take the limit...

$lim \frac{a^u-1}{u} = lim \frac{a^u ln|a|}{1} \Rightarrow ln|a|$

4. Originally Posted by colby2152
Just in case you were wondering why it is natural log, let me show you using L'Hopital's Rule.
The final substitution I left there provides a solution without L'Hôpital.