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?
 A: 
(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 $c>0$ and $f(n)\to\infty$ for every $c<0$.
A: Because $\frac{-1}{1-t}\le-1-t$,
$$
\begin{align}
\frac{\log(1-x)}x
&=\frac1x\int_0^x\frac{-1}{1-t}\,\mathrm{d}t\\
&\le\frac1x\int_0^x(-1-t)\,\mathrm{d}t\\[3pt]
&=-1-\frac{x}2
\end{align}
$$
Thus,
$$
\left(1-x\right)^{1/x}\le e^{-1-x/2}
$$
Therefore, for $c\gt0$,
$$
\begin{align}
e^{n^{3/4}}\left(\left(1-c/n^{1/4}\right)^{n^{1/4}}\right)^{n^{3/4}/c}
&=e^{n^{3/4}}\left(\left(1-c/n^{1/4}\right)^{n^{1/4}/c}\right)^{n^{3/4}}\\
&\le e^{n^{3/4}}\left(e^{-1-\frac{c}{2n^{1/4}}}\right)^{n^{3/4}}\\[3pt]
&=e^{-\frac{cn^{1/2}}2}
\end{align}
$$
A: If we set $  h=\frac{1}{n^{1/4}}   $ then 
\begin{split}\lim_{n\to \infty}f(n)&=&\lim_{n\to \infty}e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}\\
&=& \lim_{n\to \infty}e^{n^{3/4}} (1- c/n^{1/4})^{n/c}
\\&=&\lim_{h\to 0}e^{ \frac{1}{  h^3}}\left( 1-ch\right)^{1/ch^4} 
\\&=&\lim_{h\to 0}\exp\left( \frac{1}{  h^3}\right)\exp\left(     \frac{\ln(1-ch)}{ch^4} \right)\\
&=&\lim_{h\to 0}\exp\left( \frac{1}{  h^3}  \left(   \frac{\ln(1-ch)}{ch}+1\right)  \right)
\\&=& \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)
\\&=&\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 
$$\ln(1-ch)= -ch-\frac{c^2h^2}{2} -\frac{c^3h^3}{3} -\frac{c^4h^4}{4} +
o(h^4).$$
