Let's call the original principal $P$, and the interest rate $r$. Then the interest accrued in the first year is $Pr$.
If the interest is simple, the interest accrued in the second year is the same, $Pr$ again, for a total of $2Pr$. But if the interest is compound, the interest in the second year is on the original principal plus the first year's interest, $P + Pr$, and so is $(P + Pr)r = Pr + Pr^2$, rather than just $Pr$. The difference between the simple and compound interest accrued in two years is therefore $Pr^2$.
We are given that the simple interest is $2Pr = \$2,880$, and the difference between the two kinds of interest is $Pr^2 = \$160$. Dividing the second by the first gives:
$${ Pr^2\over 2Pr} = {160\over 2880} \\
\frac r2 = \frac1{18}\\
r = \frac19 $$
Or if you prefer, 11.1%.
Then we can solve for the original principal $P$, and then check the values for $P$ and $r$ by calculating the simple and compound interest amounts on $P$ at rate $r$ to see if they match the givens in the question,
Let's see. The limit claim is pretty widely discussed on MSE, so I'm assuming you're willing to believe that the limit does approach $e$. Once you have that, you can look at the formula
$$
A_n(t) = P \left(1+ \frac{r}{n}\right)^{nt}
$$
and do a little fiddling. Let $m = \dfrac{n}{r}$, so that $n = rm$. Then rewrite:
$$
A_n(t) = P \left(1+ \frac{r}{n}\right)^{nt} = P \left(1 + \frac{1}{m}\right)^{mrt}= P \left(\left(1 + \frac{1}{m}\right)^m \right)^{rt}
$$
Now as $n \to \infty$, the thing inside the large parentheses approaches $e$, so you get
$$
A(t) = Pe^{rt}.
$$
As for the main limit, the usual approach is to say you want to find
$$
L = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n
$$
but instead, you compute
\begin{align}
\ln L
&= \lim_{n \to \infty} \ln \left(1 + \frac{1}{n} \right)^n \\
&= \lim_{n \to \infty} n \ln \left(1 + \frac{1}{n} \right) \\
&= \lim_{n \to \infty} \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n},
\end{align}
which you can evaluate with L'hopital's rule (take derivative of top and bottom, since both
go towards 0):
\begin{align}
\ln L
&= \lim_{n \to \infty} \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n} \\
&= \lim_{n \to \infty} \frac{\frac{1}{1 + \frac{1}{n}}\left(\frac{-1}{n^2}\right)} {\frac{-1}{n^2}}\\
&= \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}}\\
&= 1.
\end{align}
Since the natural log of your limit is $1$, the limit itself must be $e$.$$$$$$
Best Answer
The first thing we do, let's kill all the percentages. Calculations like this are much easier when interest rates are given as pure numbers, for example $0.02$ instead of $2$%.
So if you start with principal $P$ and leave it invested at a rate $r$ (where $r = \frac R{100}$) for three years at simple interest, you end up with $P(1 + 3r)$ at the end.
But if you earn compound interest, the amount at the end is $P(1 + r)^3$.
Now take the difference in the outcomes:
$$\begin{eqnarray} P(1 + r)^3 - P(1 + 3r) &=& P(1 + 3r + 3r^2 + r^3) - P(1 + 3r) \\ &=& P(3r^2 + r^3) \\ &=& P(3 + r)r^2. \end{eqnarray}$$
Now that we know the answer, we can put it back in terms of percentages if we must:
$$P(3 + r)r^2 = P \left(3 + \frac{R}{100}\right) \left(\frac{R}{100}\right)^2.$$