# Limit as n tends to infinity of...

• March 2nd 2009, 07:53 AM
Rudipoo
Limit as n tends to infinity of...
Q: Find lim as n tends to infinity of

n(a^(1/n)-1)

where a>0.

Could anyone give me a hint of where to start please?

Thanks.
• March 2nd 2009, 08:03 AM
Plato
Rewrite as $\frac{{a^{\frac{1}{n}} - 1}}{{\frac{1}{n}}}$ then proceed.
• March 2nd 2009, 01:15 PM
redsoxfan325
Let $f(x)=a^x$ then $\lim_{n\to\infty}\frac{a^{\frac{1}{n}}-1}{\frac{1}{n}}=\lim_{x\to 0}\frac{a^x-a^0}{x-0}=f'(0)$.
In general, $f'(x)=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}=a^x\lim_{h\to 0}\frac{a^h-1}{h}$. Now let $t=a^h-1\Rightarrow h=\log_a(t+1)=\frac{\log(t+1)}{\log a}$ and when $h\to 0, t\to 0$ so $\lim_{h\to 0}\frac{a^h-1}{h}=\lim_{t\to 0}\frac{\log a}{\frac{1}{t}\log(t+1)}=\log a$.
Finally, $f'(x)=a^x\log a$.