Limit of $\frac{1}{e^3}$ I am working with a 12 year old learning about limits, and am having a tough time proving 
lim$_{n \rightarrow \infty}$ (1 - $\frac{3}{n-2})^n$ = $\frac{1}{e^3}$
using 
lim$_{n \rightarrow \infty}$ (1 + $\frac{x}{n})^n$ = lim$_{n \rightarrow 0}$ (1 + $n$)$^{\frac{x}{n}}$ = $e^x$. 
I am probably missing some small analysis trick. 
one thing I tried is that the negative 3 is why its $e^{-3}$, but where does the $n-2$ come from, how can I account for that?
ALSO, need helps showing
lim$_{n \rightarrow \infty}$ ($\frac{n+2}{n-3})^n$ = $e^5$
would you rewrite $\frac{n+2}{n-3}$ = $\frac{n}{n-3}$ + $\frac{2}{n-3}$??
 A: For the first one, take logarithms and note that
$$\lim_{n \to \infty} n \log(1 - \frac{3}{n-2}) = \lim_{n \to \infty} \frac{n}{n-2} \cdot \lim_{n \to \infty} (n-2) \log(1 - \frac{3}{n-2}).$$
The second limit on the right-hand side can be computed using your provided fact.

For the other question, Arthur has already given you a hint in the comments.
A: You could rewrite the limit:
$$\lim_{n \to \infty}\left(1-\frac{3}{n-2}\right)^n = \lim_{n \to \infty}\left(1+\frac{-3}{n-2}\right)^n = \lim_{n \to \infty}\left[\color{blue}{\left(1+\frac{-3}{n-2}\right)^{n-2}}\right]^{\color{green}{\frac{n}{n-2}}}$$
Now, just make use of the standard limit $\color{blue}{e^x = \lim_\limits{n \to \infty}\left(1+\frac{x}{n}\right)^n}$. Since $\color{green}{\frac{n}{n-2} \to 1}$ as $n \to \infty$, the limit becomes $\left(\color{blue}{e^{-3}}\right)^{\color{green}{1}} = e^{-3} = \frac{1}{e^3}$.
For your second question, I think it would be the simplest to write $\frac{n+2}{n-3}$ as $1+\frac{5}{n-3}$. The limit is therefore rewritten:
$$\lim_{n \to \infty}\left(\frac{n+2}{n-3}\right)^n = \lim_{n \to \infty}\left(1+\frac{5}{n-3}\right)^n = \lim_{n \to \infty}\left[\color{blue}{\left(1+\frac{5}{n-3}\right)^{n-3}}\right]^{\color{green}{\frac{n}{n-3}}}$$
The exact same idea applies here.
A: Just take $m=n-2$ and $x=-3$. Then you have $\lim_{m\rightarrow \infty}(1+\frac x m)^{m+2}$. You can then write that as  $\lim_{m\rightarrow \infty}(1+\frac x m)^{m}(1+\frac x m)^2$. If you can show that you can split that into separate limits, then you have ($\lim_{m\rightarrow \infty}(1+\frac x m)^{m}$) ($\lim_{m\rightarrow \infty}(1+\frac x m)^{2}$). The right parentheses clearly goes to $1$, and the left parentheses clearly goes to $e^x$. Substitute $-3$ in for $x$, and QED.

For your second limit, take $m=n-3$. Then $\frac {n+2}{n-3}=\frac {m+5} m=1+\frac5 m$. Proceed similarly.

Also, if it helps you get a more intuitive idea why $\frac {m+5} {m}$ leads to a limit of $e^5$, consider that the statement $\lim_{n\rightarrow \infty}(1+\frac 1 n )^{n}=e$ can be written as $\lim_{n\rightarrow \infty}(\frac {n+1} n)^{n}=e$. Then $\lim_{n\rightarrow \infty}(\frac {n+2}{n} )^{n}=\lim_{n\rightarrow \infty}\left(\frac {n+2}{n+1} \cdot \frac {n+1}{n}\right)^{n}$. Both $\frac{n+2}{n+1}$ and $\frac {n+1}{n}$ lead to a limit of $e$, so together they yield a limit of $e^2$. Similiary, $\frac{n+5}n$ yields a limit of $e^5$.
