Is $\lim_{n \to \infty} \left(1+\dfrac {x}{n} \right)^{n-1} = e^x$ We know that $e^x = \lim_{n \to \infty} \left(1+\dfrac {x}{n} \right)^{n}$. I was curious if we can interchange derivatives and sequence limits in this case, so I wanted to check if 
$$\dfrac {d}{dx}\lim_{n \to \infty} \left(1+\dfrac {x}{n} \right)^{n} = \lim_{n \to \infty} \dfrac{d}{dx}\left(1+\dfrac {x}{n} \right)^{n}$$
This of course turns to $$e^x = \lim_{n \to \infty} \left(1+\dfrac {x}{n} \right)^{n-1}$$
But I am not sure how to check it. Probably L'Hopitals rule would work but it would be nicer if there were another argument.
 A: In general, no, you can't interchange limits and derivatives. That being said, it is true that $e^x = \lim_{n \to \infty} \left(1+\frac {x}{n} \right)^{n-1}$. The easiest proof of that is to note that if you multiply the term inside the limit by $1+\frac xn$, you get the standard limit for $e$. And, as opposed to derivatives, limits do interact nicely with products.
A: I think this is covered by Theorem 9.18 here.
In case the link stops working, the required result is:
"Suppose that $\left(f_n\right)$ is a sequence of differentiable functions $f_n : \left(a,b\right) \rightarrow \mathbb{R}$ such that $f_n \rightarrow f$ pointwise and $f'_n \rightarrow g$ uniformly for some $f,g:
\left(a, b\right) \rightarrow \mathbb{R}$. Then $f$ is differentiable on $\left(a, b\right)$ and $f' = g$."
A: As suggested by Arthur, it suffices to observe that
$$ \left(1+\dfrac {x}{n} \right)^{n-1}=\frac{\left(1+\dfrac {x}{n} \right)^{n}}{\left(1+\dfrac {x}{n} \right)^{}} \to \frac{e^x}{1}=e^x$$
A: Yes its True.
This limit is of $1^ \infty $ form.  
Take $$ lim_{n \to \infty} (1+ \frac{x}{n})^{n-1}=L$$
Take ln on both sides:
$$ln(L)= \lim_{{n \to \infty}}(n-1) \cdot ln(1+ \frac{x}{n})$$
Approximate as:
$$ ln(L) =  \lim_{{n \to \infty}}(n-1) (1+ \frac{x}{n} -1) $$
$$ ln(L) =  \lim_{{n \to \infty}}(\frac{x \cdot (n-1)}{n}) $$
$$ln(L)=x$$
$$L=e^x$$
