<s>[INST] Find $\lim_{n\rightarrow\infty} \left(\frac 1e\left( 1+\frac1n+\frac c{n^2}\right)^n\right)^n$ 
Find
  $$\lim_{n\rightarrow\infty} \left(\frac 1e\left( 1+\frac1n+\frac c{n^2}\right)^n\right)^n.$$

According to Wolfram the limit is $e^{c-1/2}$. I have simplified the expression to
$$ \exp \lim \left[ n^2 \ln \left(1 + \frac{1}{n} + \frac{c}{n^2} \right) - n\right]. $$
I want to approximate $ \ln \left(1 + \frac{1}{n} + \frac{c}{n^2}\right) $ with
a Taylor polynomial with $ n=3 $, $ x_0 =0 $ and $ f(x) = \ln (1+x) $.
I have tried 
$$ f(1/n) = \frac{1}{n} - \frac{\left(\frac{1}{n}\right)^2}{2} +
\frac{\left(\frac{1}{n}\right)^3}{3} - b_n$$
with $ |b_n| < \frac{1}{4n^4} $.
Then
$$ e^{\lim\left[ n - \frac{1}{2} + \frac{c}{3n} + b_n n^2 - n \right] }.$$
So $ b_n n^2 $ will go to 0 and the $ n $'s will cancel out.
I'm not sure how to get $ -\frac{1}{2} + c $ in the exponent because of the $ 3n $ in the 
denominator.
 [/INST] Put $x=\dfrac 1n+\dfrac c{n^2}$, then
\begin{align*}
n^2f(x)-n&=\color{blue}{n^2}\left(\dfrac 1n+\color{blue}{\dfrac c{n^2}}-\frac12\left(\dfrac 1n+\dfrac c{n^2}\right)^2+o(\frac1{n^2})\right)-n\\
&=\color{blue}c-\frac12+o(1).
\end{align*}
</s>