# $\lim_{n\rightarrow\infty}(1+1/a_n)^n=\lim e^{n/a_n}$, where $\lim_{n\rightarrow\infty}{a_n}=\infty$

We know that $$\lim_{n\rightarrow\infty}(1+1/n)^n = e$$
The following that $$\lim_{n\rightarrow\infty}(1+1/a_n)^n=\lim e^{n/a_n}$$,
where $$a_n$$ is positive sequence and $$\lim_{n\rightarrow\infty}{a_n}=\infty$$ is true?

• More generally, $$\lim A_n^{B_n} = (\lim A_n)^{\lim B_n}$$ unless it is one of the usual indeterminate forms like $0^0$ or $1^\infty$. – GEdgar Jan 24 at 14:42
• Thank you for asking my questions.~ – Hs P Jan 24 at 15:53

Yes, that is true. Hint: You can show that $$\underset{x\to a}{\lim}{\ f(x)}=1$$ and $$\underset{x\to a}{\lim}{\ g(x)}=+\infty$$ implies that $$\underset{x\to a}{\lim}{\ f(x)^{g(x)}}=e^{\underset{x\to a}{\lim}{\ \left(f(x)-1\right)g(x)}}$$ whenever $$\underset{x\to a}{\lim}{\ \left(f(x)-1\right)g(x)}$$ exists.
To see this you can Write the following: $$f(x)^{g(x)}=\left(\left(1+(f(x)-1)\right)^{\displaystyle{\frac{1}{f(x)-1}}}\right)^{(f(x)-1)g(x)}$$ then you can use the assertion of GEdgar's comment. You have to recall that
$$\underset{u\to 0}{\lim}{\ (1+u)^{\displaystyle{\frac{1}{u}}}} = e$$
$$\underset{x\to a }{\lim}{\ \bigg(1+(f(x)-1)\bigg)^{\displaystyle{\frac{1}{f(x)-1}}}} = \underset{u\to 0}{\lim}{\ (1+u)^{\displaystyle{\frac{1}{u}}}}=e$$ using the change of variables $$u=f(x)-1$$. Then, $$u\to 0$$ as $$x\to a$$.