Easy limit calculation $\lim_{n\to +\infty} n(a^{1/n} -1)$

Question

I have to calculate the limit $\lim_{n\to +\infty} n(a^{1/n} -1)$.

I found that it tends to $a$ but don't really see how to prove it with one or 2 steps... Can you please help me out ?

score 3 · Accepted Answer · answered Oct 14 '14 at 16:18

3

Set $\dfrac1x=h$

$$F=\lim_{n\to\infty}n(a^{\frac1n}-1)=\lim_{h\to0}\frac{a^h-1}h$$

Now, $a=e^{\ln a}\implies a^h=(e^{\ln a})^h=e^{h\ln a}$

$$\implies F=\ln a\lim_{h\to0}\frac{e^{h\ln a}-1}{h\ln a}=?$$

answered Oct 14 '14 at 16:18

lab bhattacharjee

still doesnt take me anywhere – C. Chris Oct 14 '14 at 16:34
@C.Chris, Do you know http://math.stackexchange.com/questions/152605/proving-that-lim-limits-x-to-0-fracex-1x-1 – lab bhattacharjee Oct 14 '14 at 16:36
Got it thanks :) – C. Chris Oct 14 '14 at 16:54

score 0 · Answer 2 · answered Oct 14 '14 at 16:27

0

If $f(x)=a^x$, then $$ \log a=f'(0)=\lim_{h\to 0}\frac{a^h-1}{h}=\lim_{n\to \infty}\frac{a^{1/n}-1}{1/n} =\lim_{n\to \infty}n(a^{1/n}-1). $$

answered Oct 14 '14 at 16:27

2 Answers2