provided $f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}$, what is f(x)? $$
f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}
$$
Here is one of the solution from my workbook:
$$
 \begin{array}{l}
f( x+1) \ =\ \lim _{n\rightarrow \infty }\left[\left( 1+\frac{2+x}{n-2}\right)^{\frac{n-2}{2+x}}\right]^{n\left(\frac{2+x}{n-2}\right)} =\ e^{2+x} =e^{1+( x+1)}\\
f( x) \ =\ e^{1+x}
\end{array}
$$
but is a bit preplexing for me :(
 A: First, $$\frac{n+x}{n-2} = 1 + \frac{1}{\frac{n-2}{2+x}}$$
from rearranging.
Then they use the following fact $$\lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^k = e$$
which implies
$$\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x}} = e.$$
By continuity, we can raise both sides to the $2+x$ power.
$$\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x} \cdot (2+x)} = e^{2+x}.$$
Finally, note that
$$\begin{align}
\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x} \cdot \frac{n}{n-2}(2+x)}
&=
\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x} \cdot (2+x)}
\cdot
\underbrace{\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{2}}_{=1}
\\
&=
\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x} \cdot (2+x)}
\\
&= e^{2+x}.\end{align}$$
A: $$y=\left(\frac{n+x}{n-2}\right)^{n}=\left(\frac{(n-2)+(x+2)}{n-2}\right)^{n}=\left(1+\frac{x+2}{n-2}\right)^{n}$$ Taking logarithms
$$\log(y)=n \log\left(1+\frac{x+2}{n-2}\right)\sim n \frac{x+2}{n-2}=\frac{n}{n-2}(x+2)\sim x+2$$ So
$$y=f(x+1)\sim e^{x+2} \implies f(x)=e^{x+1}$$
