Please help me evaluate this messy limit

Question

I'd like to solve a limit of this form: $$\lim_{n\rightarrow\infty} \frac{1}{n}\Big[\log\Big( \frac{\sqrt{2\pi n} (x-p) - p(1-p) e^{-\frac{n}{2}(\frac{x-p}{p(1-p)})^2}}{\sqrt{2\pi n} (x-p)} \Big)\Big] $$ But I'm not sure how to start? Does anyone have any idea - can we use L'Hopital's rule here?

Answer 1

It's not an indeterminate form so we don't have to use anything other than basic limit laws. Also, we assume $x\ne p$ and $p(p-1)\ne 0$ for the divisions to be defined. $$\log\left( \frac{\sqrt{2\pi n} (x-p) - p(1-p) e^{-\frac{n}{2}\left(\frac{x-p}{p(1-p)}\right)^2}}{\sqrt{2\pi n} (x-p)} \right)=\log\left(1-\frac{p(1-p) e^{-\frac{n}{2}\left(\frac{x-p}{p(1-p)}\right)^2}}{\sqrt{2\pi n} (x-p)}\right) $$ Notice that $$e^{-\frac{n}{2}\left(\frac{x-p}{p(1-p)}\right)^2}\to 0,\frac{1}{\sqrt{2\pi n} (x-p)}\to 0\,\,(n\to\infty) $$ so the logarithm approaches $\log(1)=0$. And $\frac 1n\to 0$. Therefore, the limit is $0$.

Answer 2

$$\lim_{n\rightarrow\infty} \frac{1}{n}\Big[\log\Big( \frac{\sqrt{2\pi n} (x-p) - p(1-p) e^{-\frac{n}{2}(\frac{x-p}{p(1-p)})^2}}{\sqrt{2\pi n} (x-p)} \Big)\Big] $$ $$=\lim_{n\rightarrow\infty} \frac{1}{n}\Big[\log\Big(1- \frac{ p(1-p) e^{-\frac{n}{2}(\frac{x-p}{p(1-p)})^2}}{\sqrt{2\pi n} (x-p)} \Big)\Big] $$ $$=\lim_{n\rightarrow\infty} \frac{1}{n}\Big[- \frac{ p(1-p) e^{-\frac{n}{2}(\frac{x-p}{p(1-p)})^2}}{\sqrt{2\pi n} (x-p)} +O(\frac{e^{-n(\frac{x-p}{p(1-p)})^2}}{n})\Big] $$ $$=\lim_{n\rightarrow\infty} - \frac{ p(1-p) e^{-\frac{n}{2}(\frac{x-p}{p(1-p)})^2}}{\sqrt{2\pi n^3} (x-p)} +O(\frac{e^{-n(\frac{x-p}{p(1-p)})^2}}{n^2}) $$ $$=-\lim_{n\rightarrow\infty} p(1-p) e^{-\frac{n}{2}(\frac{x-p}{p(1-p)})^2} \cdot \lim_{n\rightarrow\infty}\frac{ 1}{\sqrt{2\pi n^3} (x-p)} =0 $$