Determining $\ {\displaystyle\lim_{n\to\infty}}\left(\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}\right)$ This limit was a trivia question.  The trivia league is set up so you have all day to mull over questions and submit your answer, but it's not "trivia for mathematicians."  The questions tend to be tough, but I wouldn't expect them to ask a really really hard Calculus problem...

$$\lim_{n\to\infty}\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}$$

I can provide the correct answer but it's easy to guess by plugging in a large number for $n$.  No calculators allowed!
 A: 
The evaluation of the limit requires consideration of more than simply the limit definition of $e$.  Note that we can write $$\frac{(n+1)^{n+1}}{n^n}-\frac{n^{n}}{(n-1)^{n-1}}=(n+1)\left(1+\frac1n\right)^n-(n-1)\left(1-\frac1n\right)^{-n}$$
If we then proceed naively and incorrectly by suggesting that this is "equivalent" to $(n+1)e-(n-1)e=2e$, we obtain the wrong answer.  We need to consider more carefully higher-order terms to arrive at the correct answer.  To that end, we now proceed. 


We will use the expansions $\log(1\pm x)=x\mp \frac12 x^2+O(x^3)$ and $e^x=1+x+\frac12x^2$ in the following development.
Using the aforementioned expansions, we can write
$$\begin{align}\frac{(n+1)^{n+1}}{n^n}&=(n+1)\left(1+\frac{1}{n}\right)^n\\\\
&=(n+1)e^{n\log\left(1+\frac{1}{n}\right)}\\\\
&=(n+1)e\left(1-\frac1{2n}+O\left(\frac{1}{n^2}\right)\right)\tag 1
\end{align}$$
and
$$\begin{align}
\frac{n^{n}}{(n-1)^{n-1}}&=(n-1)\left(1-\frac{1}{n}\right)^{-n}\\\\
&=(n-1)e^{-n\log\left(1-\frac{1}{n}\right)}\\\\
&=(n-1)e\left(1+\frac1{2n}+O\left(\frac{1}{n^2}\right)\right)\tag2
\end{align}$$

Subtracting $(2)$ from $(1)$ reveals
$$\begin{align}
\frac{(n+1)^{n+1}}{n^n}-\frac{n^{n}}{(n-1)^{n-1}}&=e+O\left(\frac1n\right)\\\\
\end{align}$$
whence taking the limit as $n\to \infty$ yields the coveted limit
$$\lim_{n\to \infty}\left(\frac{(n+1)^{n+1}}{n^n}-\frac{n^{n}}{(n-1)^{n-1}}\right)=e$$
