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I know that $ e^q = \lim\limits_{n \rightarrow \infty} \sum_{k=0}^n \frac{q^k}{k!}$ and $\lim\limits_{n \rightarrow \infty}{(1+ \frac{1}{n})^n}=e$ .

But what is the general limit of $(1+ \frac{x}{n^p})^n$ and how can i prove it by using these equantions above?

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Note that, if $a_n$ is a sequence that satisfies $\lim_{n\to +\infty} a_n=+\infty$, then $$\lim_{n\to +\infty} \left(1+\frac{1}{a_n}\right)^{a_n}=e$$

Using this, we have, for $x\in\mathbb R$ fixed, $$\lim_{n\to+\infty} \left(1+\frac{x}{n^p} \right)^n=\lim_{n\to+\infty} \left(1+\frac{1}{\frac{n^p}{x}} \right)^n=\lim_{n\to+\infty} \left[\left(1+\frac{1}{\frac{n^p}{x}} \right)^{\frac{n^p}{x}}\right]^\frac{x}{n^{p-1}}=e^L$$

where $L=\lim_{n\to +\infty} \frac{x}{n^{p-1}}$.

For the particular case of $p=1$, $\lim_{n\to+\infty} \left(1+\frac{x}{n} \right)^n=e^x$.

If you need further explanation, please ask.

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    Just to add on to the (very good) answer of flytothesurface, in general you will have $\lim_{n \to \infty}(1 + \frac{x}{n})^n = e^x$. It is in fact possible to start with this limit as the *definition* of $e^x$, and show in an elementary way that this function will have the familiar properties of $e^x$ (without making reference to the series expansion for $e^x$). Perhaps the trickiest part is showing from first principles that $\lim_{n \to \infty}(1 + \frac{x}{n})^n$ converges for all real $x$.... – 2017-02-11
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    @flytothesurface Thank you! I just have one question. Why is this equation $( 1+ \frac{1}{a_n} )^{a_n} = e$ true as n goes to infinity? – 2017-02-11
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    @Valentin it is only true if $\lim_{n\to +\infty} a_n =+\infty$. Intuitively, you can think that if $a_n\to +\infty$, taking $\lim_{n\to +\infty} \left(1+\frac{1}{a_n}\right)^{a_n}$ would make no difference, as $a_n$ behaves similarly to the sequence $b_n=n$. – 2017-02-11
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    More rigorously, if $f$ is continuous, and $\lim_{n\to +\infty} a_n=L$ then $\lim_{n\to +\infty} f(a_n)=f(L)$. If $L=+\infty$, a similar equality holds: $$\lim_{n\to +\infty} f(a_n)=\lim_{x\to +\infty} f(x)$$ In this case, take $f(x)=\left(1+\frac1x\right)^x$, which satisfies $\lim_{x\to +\infty} f(x)=e$ – 2017-02-11
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    @flytothesurface Thanks alot :) – 2017-02-11
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    Just note that the statement $$\lim_{n\to\infty} \left(1+\dfrac{1}{n}\right)^{n} =e\tag{1}$$ assumes that $n$ is a positive integer (the statement is true even if $n$ is a real variable but then the equation is on a different level of complexity and is not an immediate consequence of equation $(1)$). Hence we should add that if $a_{n} $ is a sequence which takes *positive integer values* and if $a_{n} \to\infty$ as $n\to\infty$ then $(1+(1/a_{n}))^{a_{n}}\to e$ as $n\to\infty$. Cont'd... – 2017-02-12
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    If $s_{n} =(1+(1/n))^{n}$ then the sequence $t_{n} =s_{a_{n}} $ is a subsequence of $s_{n} $ and hence it also converges to the same limit $e$. – 2017-02-12