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?

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    $\begingroup$ 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 $1^\infty$. $\endgroup$ – GEdgar Jan 24 at 14:42
  • $\begingroup$ Thank you for asking my questions.~ $\endgroup$ – 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

$$\underset{u\to 0}{\lim}{\ (1+u)^{\displaystyle{\frac{1}{u}}}} = e $$

and, so we get

$$ \underset{x\to a }{\lim}{\ \bigg(1+(f(x)-1)\bigg)^{\displaystyle{\frac{1}{f(x)-1}}}} = \underset{u\to 0}{\lim}{\ (1+u)^{\displaystyle{\frac{1}{u}}}}=e $$ using the change of variables $u=f(x)-1$. Then, $u\to 0$ as $x\to a$.

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    $\begingroup$ Thank you for asking my questions. I'll dive into your answer~!! $\endgroup$ – Hs P Jan 24 at 15:54

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