I am trying to find the limit (If it does exist)

$\lim_{n\rightarrow\infty}\left(1-|\mathcal{X}|^{-\alpha n}\right)^{2^{nC}\left(1-|\mathcal{X}|^{-\alpha n}\right)}$, where $0<\alpha<1$, $C>0$, and $|\mathcal{X}|\geq 2$.

And, in case that it does not exist in general, can we find extra conditions on $\alpha, C, |\mathcal{X}|$ that makes the limit exist?

I have tried the following:

Let $f(n)=\left(1-|\mathcal{X}|^{-\alpha n}\right)$, $g(n)=2^{nC} f(n)$, and $h(n)=\frac{1}{g(n)}$. Now, we need to find the limit $\lim_{n\rightarrow\infty} f(n)^{g(n)}$. So, I proceeded as follows: \begin{align} \lim_{n\rightarrow\infty} f(n)^{g(n)}&=\lim_{n\rightarrow\infty} g(n) \ln f(n)\\ &=\lim_{n\rightarrow\infty} \frac{\ln f(n)}{h(n)}\\ &=\frac{0}{0}. \end{align}

Then, I have tried to use L'hopital rule by computing $\frac{d}{dn}\log f(n)$ and $\frac{d}{dn}h(n)$ and finding $\lim_{n\rightarrow\infty} \frac{\frac{d}{dn}\log f(n)}{\frac{d}{dn}h(n)}$, but it is equal to $\frac{0}{0}$ as well! Does that mean that the limit does not exist? And if yes, can it exist for some specific values of $\alpha, C, |\mathcal{X}|$?


Notice that $$\ln\left(1-|\mathcal{X}|^{-\alpha n}\right)^{2^{nC}\left(1-|\mathcal{X}|^{-\alpha n}\right)}=2^{nC}\left(1-|\mathcal{X}|^{-\alpha n}\right)\ln\left(1-|\mathcal{X}|^{-an}\right).$$ Now we have to treat the cases

  1. $2^C/\mathcal X^a\gt 1$;
  2. $2^C/\mathcal X^a= 1$;
  3. $2^C/\mathcal X^a\lt 1$.
| cite | improve this answer | |
  • $\begingroup$ I can not see this rigorously, however, I understand that for convergence I need that $|\mathcal{X}|^{-n\alpha}$ goes to zero faster than $2^{nC}$ goes to infinity (i.e., $2^{C}/|\mathcal{X}|^{\alpha}<1$.) Can you explain more? $\endgroup$ – Meemo Dec 23 '14 at 0:33
  • $\begingroup$ You have to consider $\lim_{x\to 0}\ln(1-x)/x$ if you want to make this rigorous. $\endgroup$ – Davide Giraudo Dec 23 '14 at 10:01

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.