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 $e$.$$$$$$
Considering the equation$$A = M \frac{\left(1+\frac{R}{N}\right)^{NT}-1}{\frac RN}\left(1+\frac{R}{N}\right)\tag1$$ define
$$a=\frac AM \qquad, \qquad x=\frac RN\qquad,\qquad n=NT$$ to make the equation
$$a=\frac{(1+x)^n-1} x(1+x) \tag 2$$ What we know is that $x\ll 1$; so, let us develop the rhs using the binomial theorem
$$a=n+\sum_{k=1}^\infty \binom{n+1}{k+1}x^k$$ Transform it as simple Padé approximants which could be
$$a=n\frac{ (n+5) x+6}{6-2 (n-1) x}\implies x=\frac{6 (a-n)}{2 a (n-1)+n (n+5)}$$
Let us try with $a=500$ and $n=200$. The above formula would give
$$a=\frac{3}{400}=0.00750$$ while the exact solution would be $0.00809$; this is not so bad taking into account the huge value assigned to $a$.
Better, but at the price of a quadratic equation
$$a=n\frac{60 +6 (n+13) x+ (n^2+3 n+20) x^2 } {60-24 (n-2) x+3 (n-2) (n-1) x^2 }$$ For the worked example, selecting the "reasonable" root, this would give $x=0.00812$
For sure, we could continue improving but this would be at the price of cubic or quartic equations which can be solved with radicals. To give you an idea, using cubic equations, we should get $x=0.00809422$ while the exact solution is $0.00809450$.
Play with that and, please, tell me how it works for your cases.
Edit
The first way described above is "neutral" in the sense that $a$ is set equal to the ratio of polynomials leading to linear, quadratic, cubic or quartic equations in $x$.
There is another way. Rewrite $(2)$ as $(3)$
$$\frac 1a=\frac{x}{(x+1) \left((x+1)^n-1\right)}\tag 3$$ Expand the rhs as a Taylor series centered at $x=0$ and use series reversion to get
$$x=t+\frac {b_1} 6 t^2+\frac {b_2} {36} t^3+\frac {b_3} {1080} t^4+\frac {b_4} {6480} t^5+\frac {b_5} {90720} t^6+\frac {b_6} {2721600} t^7+O(t^{8})$$ where $t=\frac{2(a-n)}{(n+1)a}$. The coefficients are listed in the table below
$$\left(
\begin{array}{cc}
k & b_k \\
1 & n+5 \\
2 & 2 n^2+11 n+23 \\
3 & 22 n^3+153 n^2+402 n+503 \\
4 & 52 n^4+428 n^3+1437 n^2+2438 n+2125 \\
5 & 300 n^5+2836 n^4+11381 n^3+24879 n^2+30911 n+20413 \\
6 & 3824 n^6+40692 n^5+188712 n^4+496259 n^3+799917 n^2+780417 n+411779
\end{array}
\right)$$
Defining
$$x_{(p)}=t+\sum_{k=0}^p \frac{b_k}{c_k}\,t^{k+1}$$ for the worked example we should get the following values
$$\left(
\begin{array}{cc}
p & x_{(p)} \\
0 & 0.0059701493 \\
1 & 0.0071879409 \\
2 & 0.0076739522 \\
3 & 0.0078882746 \\
4 & 0.0079897311 \\
5 & 0.0080399577 \\
6 & 0.0080655906 \\
7 & 0.0080789649 \\
\cdots & \cdots \\
\infty &0.0080945103
\end{array}
\right)$$
Best Answer
The problem is just after $$1.48595=\left(1+\frac{r}{2}\right)^{20}$$ You have now $$\sqrt[20]{1.48595}=1+\frac{r}{2}\implies r=0.04$$