1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $x\to a$.

$\endgroup$
  • 1
    $\begingroup$ Thank you for asking my questions. I'll dive into your answer~!! $\endgroup$ – Hs P Jan 24 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.