Limit of $e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}$ Let $c\ne0$ be a constant. Consider the limit of $$f(n)=e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}$$ as $n \to \infty$. 


I think it is zero because for large $n$,

$$e^{n^{3/4}} ((1- \frac{c}{n^{1/4}})^{n^{1/4}})^{\frac{n^{3/4}}{c}} \approx e^{n^{3/4}} (e^{-c})^{ \frac{n^{3/4}}{c}}$$

But, how do I prove it formally?
(2017-10-24) Amusing downvote, purely for mathematical reasons, I am sure...


Thus, using the expansion $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$, one sees that $$\log f(n)=n^{3/4}+c^{-1}n\log(1-cn^{-1/4})$$ is also $$\log f(n)=n^{3/4}+c^{-1}n\,(-cn^{-1/4}-\tfrac12c^2n^{-1/2}+o(n^{-1/2}))$$ that is, $$\log f(n)=-\tfrac12cn^{1/2}+o(n^{1/2})$$ In particular, $f(n)\to0$ for every $c&gt;0$ and $f(n)\to\infty$ for every $c&lt;0$.
If we set $  h=\frac{1}{n^{1/4}}   $ then 

\begin{split}\lim_{n\to \infty}f(n)&amp;=&amp;\lim_{n\to \infty}e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}\\
&amp;=&amp; \lim_{n\to \infty}e^{n^{3/4}} (1- c/n^{1/4})^{n/c}
\\&amp;=&amp;\lim_{h\to 0}e^{ \frac{1}{  h^3}}\left( 1-ch\right)^{1/ch^4} 
\\&amp;=&amp;\lim_{h\to 0}\exp\left( \frac{1}{  h^3}\right)\exp\left(     \frac{\ln(1-ch)}{ch^4} \right)\\
&amp;=&amp;\lim_{h\to 0}\exp\left( \frac{1}{  h^3}  \left(   \frac{\ln(1-ch)}{ch}+1\right)  \right)
\\&amp;=&amp; \lim_{h\to 0}\exp\left( \frac{1}{  h^3}  \left(   -\frac{ch } {2}  -\frac{c^2h^2}{3} -\frac{c^3h^3}{4}
 +o(h^3)\right)  \right)
\\&amp;=&amp;\lim_{h\to 0}\exp\left( -\frac{c } {2  h^2}  -\frac{c^2}{3h} -\frac{c^3}{4}
 +o(1)\right)=0\end{split}

Given that 
$$\ln(1-ch)= -ch-\frac{c^2h^2}{2} -\frac{c^3h^3}{3} -\frac{c^4h^4}{4} +
o(h^4).$$