# Limit of a function

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. $2^C/\mathcal X^a\lt 1$.
• 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? – Meemo Dec 23 '14 at 0:33
• You have to consider $\lim_{x\to 0}\ln(1-x)/x$ if you want to make this rigorous. – Davide Giraudo Dec 23 '14 at 10:01