TITLE: Is $\lim_{n \to \infty} \left(1+\dfrac {x}{n} \right)^{n-1} = e^x$
QUESTION [2 upvotes]: 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.

REPLY [3 votes]: As suggested by Arthur, it suffices to observe that
$$ \left(1+\dfrac {x}{n} \right)^{n-1}=\frac{\left(1+\dfrac {x}{n} \right)^{n}}{\left(1+\dfrac {x}{n} \right)^{}} \to \frac{e^x}{1}=e^x$$

REPLY [1 votes]: 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:
$$ ln(L) =  \lim_{{n \to \infty}}(n-1) (1+ \frac{x}{n} -1) $$
$$ ln(L) =  \lim_{{n \to \infty}}(\frac{x \cdot (n-1)}{n}) $$
$$ln(L)=x$$
$$L=e^x$$
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