Expressing $e^{f(x)}$ as the limit as $x$ goes to infinity of a function. Can someone help me prove that the function $(1+f(a)/x)^x$ as $x $ goes to infinity is equal to $e^{f(a)}$??
I do understand that e = $1+1+(1/2)+(1/3!)+\cdots$
and in general that $e^x = 1+x+(x/2)+(x/3!)+\cdots$
Thank u!!!
 A: So there is something wrong with your limit (once the limit is taken the limit should not depend on $x$). Thus I suspect that the variable inside $f$ is not the limit variable. First note that:
\begin{align*}
 e^x = \lim_{t \to \infty} \left(1+ \frac{x}{t} \right)^t.
\end{align*}
Hence we have, that:
\begin{align*}
 e^{f(x)} = \lim_{t \to \infty} \left(1+ \frac{f(x)}{t} \right)^t
\end{align*}
A: Another way to do it : consider $$y=\left(1+\frac{f(a)}{x}\right)^x$$ Take logarithms $$\log(y)=x \log\left(1+\frac{f(a)}{x}\right)$$ Since $x\to \infty$, $\frac{f(a)}{x}$ is small. Use now the fact that, using equivalents $\log(1+\epsilon)\sim \epsilon$. Replace $\epsilon$ by $\frac{f(a)}{x}$.
So $$\log(y)\sim x \cdot \frac{f(a)}{x}=f(a)$$ Now, using $y=e^{\log(y)}$ then, for large $x$ $$y\sim e^{f(a)}$$
If you prefer to use Taylor series, since, around $\epsilon=0$ $$\log(1+\epsilon)=\epsilon -\frac{\epsilon ^2}{2}+O\left(\epsilon ^3\right)$$ then $$\log(y)=x \left(\frac{f(a)}{x}-\frac{f(a)^2}{2 x^2}+O\left(\frac{1}{x^3}\right)\right)=f(a)-\frac{f(a)^2}{2 x}+O\left(\frac{1}{x^2}\right)$$ Taylor again leads to $$y=e^{f(a)}-\frac{e^{f(a)} f(a)^2}{2 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.
A: A friend was able to explain it to me.
using the binomial theorem for $(1+x/n)^n$ as "$n$" goes to infinity we get that the expression equals
$( 1 + x + (n!(x/n)^2)/2!(n-2)! + (n!(x/n)^3)/3!(n-3)! + ... $
= $(1+x+((n-1)nx^2)/2!n+((n-2)(n-1)nx^3)/3!n^3 + ...$
as "$n$" goes to infinity we get
= $(1 + x + x/2! + x/3! +...)$
which is equal to $e^x$ (taylor expansion)
And so, since $e^x = (1+x/n)^n$ as "$n$" goes to infinity we can represent $(1+f(x)/n)^n$ as $(1+y/n)^n$ which equals $e^y$ and we are done.
Thank you everyone who responded.
