I have this equation:
$$\lim_{n\to \infty}\left(1+{x\over n} \right)^n=e^x$$
I need to Taylor expand both sides to prove that they are equal. Now I know how to do this using l'Hopital's rule, but I'm not supposed to prove this in that way. This is what I've tried so far:
$$\lim_{n\to \infty}\left(1+{x\over n} \right)^n$$ $$=\lim_{n\to \infty}e^{n\big[\log\big( 1+{x\over n}\big)\big]}$$ $$=\lim_{n\to \infty}\big(1+n\log\big(1+{x\over n}\big)+{n^2\over 2!}{\log^2\big(1+{x\over n}}\big)+O(x^3)\big)$$
This doesn't look anything like the expansion for $e^x$, so I tried to pass the limit through the exponential, expand $\log_{n\to \infty}$, and solve it that way:
$$e^{\lim_{n\to \infty}\big(1+{x\over n} \big)^n}$$ $$e^{\lim_{n\to \infty}\big[1+n\big({x\over n}-{x^2\over 2n^2}+{x^3\over 3n^3}-O(x^4)\big) \big]}$$
Distributing the $n$ into the parentheses, and taking the limit of my result was this:
$$e^{\lim_{n\to \infty}\big(1+{x\over n} \big)^n} = e^{1+x}=e^1e^x$$
And it looks very close to what I need it to be. I also have the problem that there aren't any factorials in the denominators of my expansion. I found this stack exchange question: Limit Question involving logarithmic taylor expansion. However, it didn't help me very much.
How should I expand this equation and what am I doing wrong?