NOTE 1: L'Hospitals and Taylor expansions are not allowed.
NOTE 2: I really appreciate if someone would correct my attempt, however any other easier method only involving single variable calculus (excluding the concepts in NOTE 1) are welcome.
PROBLEM: Compute $$\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.$$
I'll just manipulate without writing out the limit, for now. I have
\begin{array}{lcl}
\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n} & = & \left( 1+\frac{1}{n}\right)^{n^2}\cdot e^{-n} \\
& = & \exp\left( n^2\ln\left(1+\frac{1}{n}\right)-n\right) \\
& = & \exp((n\ln(1+\frac{1}{n})-1)n) \\
\end{array}
And proceeding:
\begin{array}{lcl}
\exp((n\ln(1+\frac{1}{n})-1)n) & = & e^{((n\ln(1+\frac{1}{n})-1)n} \\
& = & (e^{((n\ln(1+\frac{1}{n})-1)})^n \\
& = & \left(\frac{(e^{((n\ln(1+\frac{1}{n})-1)}-1+1)}{(n\ln(1+\frac{1}{n})-1}\cdot{((n\ln(1+\frac{1}{n})-1})\right)^n\\
\end{array}
It gets quite ugly very quickly as you can see. I'm trying to rewrite it so I can apply standard limits like $$\lim_{x\rightarrow\infty}\frac{e^x-1}{x}=\infty \quad \text{and} \quad \lim_{x\rightarrow\infty}\frac{\ln{(1+x)}}{x}=0.$$
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