Limit $\lim (1+(a^{1/n}-1) / b)^n$

Question

How can I find the $\lim (1+(a^{1/n}-1) / b)^n$ when $n \to \infty$?

I would greatly appreciate it if you kindly give me some hints.

Answer 1

One can get the answer easily by taking logs. We have $$n\log(1 + (a^{1/n} - 1)/b) = n\frac{\log(1 + t)}{t}\cdot t$$ and hence the limit is same as that of $n(a^{1/n} - 1)/b$ which tends to $(\log a)/b = \log a^{1/b}$. Thus original limit is $a^{1/b}$. It is necessary to assume $a > 0$ for the above to work.

We have used the fact that $a^{1/n} \to 1$ so that $t = (a^{1/n} - 1)/b \to 0$ and hence $(\log(1 + t))/t \to 1$. Further use is made of the standard limit $n(a^{1/n} - 1) \to \log a$.

Answer 2

Let $N=\lim_{n\to\infty}\left(1+\dfrac{a^{1/n}-1}b\right)^n$

$\ln N=\dfrac1b\lim_{n\to\infty}\dfrac{\ln\left(1+\dfrac{a^{1/n}-1}b\right)}{\dfrac{a^{1/n}-1}b}\cdot\lim_{n\to\infty}\dfrac{a^{1/n}-1}{1/n}$

Now use $\lim_{u\to0}\dfrac{\ln(1+u)}u=1$

and $\lim_{h\to0}\dfrac{a^h-1}h=\ln a$