1)if$$e = \lim_{n\to∞}(1+\frac{1}{n})^n$$
2)then how can this be correct $$e^x = \lim_{n\to∞}(1+\frac{x}{n})^n$$
3)shouldn't it be $$e^x = \lim_{n\to∞}((1+\frac{1}{n})^n)^x$$
how can the power $x$ just get inside the definition of $e$ as in case $2$. Someone please explain to to me this
On
Indeed, (2) and (3) are the same. Let's make this more believable by rewriting (1) to this form. $$e = \lim_{n\to\infty}(1 + \frac{1}{n})^{n} = \lim_{n\to\infty}(1 + \frac{1}{n})^{\frac{n}{1}}.$$
See how in the exponent there is the inverse ($\frac{n}{1}$) of what is in the bracket $(\frac{1}{n})?$. We can also replace the $1$ for our given $x$, and the limit will still be $e$.
$$\lim_{n\to\infty}(1 + \frac{x}{n})^{\frac{n}{x}} = e.$$
However, our goal is to find $\lim_{n\to\infty}(1 + \frac{x}{n})^{n}$. So we raise the previous limit to the $x$:
$$\lim_{n\to\infty}(1 + \frac{x}{n})^{n} =\lim_{n\to\infty}((1 + \frac{x}{n})^{\frac{n}{x}})^{x} = (\lim_{n\to\infty}(1 + \frac{x}{n})^{\frac{n}{x}})^{x} = e^{x}.$$
Note that for the last step, we are taking the limit inside the $(\cdot)^{x}$, but this is allowed since $x$ is a constant relative to $n$ and the exponent is a continuous operation.
The following extensive elaboration of Abezhiko's comment may help. To help keep the variables and constants apart in our mind, I'll use the specific number $17$ instead of $x.$
$$ e^{17} \; = \; \left[ \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^n \right]^{17} \; = \; \lim_{n \rightarrow \infty} \left[ \left(1 + \frac{1}{n} \right)^n \right]^{17} \; = \; \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^{17n} $$
The 1st equality is the result of raising both sides of your (1) to the $17$th power. The 2nd equality is by continuity of the function $f(t) = t^{17}.$ The 3rd equality is by a property of exponents. Continuing $\ldots$
$$ \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^{17n} \; = \; \lim_{17n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^{17n} $$
This follows because the limit operation "$n \rightarrow \infty$" is the same as the limit operation $17 \rightarrow \infty$" (i.e. does nothing more and does nothing less). That is, $\;n \rightarrow \infty\;$ implies $\;17n \rightarrow \infty\;$ AND $\;17n \rightarrow \infty\;$ implies $\;n \rightarrow \infty.$ (Note, for example, that neither of the limit operations $n^2 \rightarrow \infty$ and $\frac{5}{n} \rightarrow \infty$ are the same as the limit operation $n \rightarrow \infty.)$ Now change variables by letting $m = 17n.$ Continuing $\ldots$
$$ \lim_{17n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^{17n} \; = \; \lim_{m \rightarrow \infty} \left(1 + \frac{1}{n} \right)^{m} \; = \; \lim_{m \rightarrow \infty} \left(1 + \frac{1}{m/17} \right)^{m} \; = \; \lim_{m \rightarrow \infty} \left(1 + \frac{17}{m} \right)^{m} $$
The 1st equality is the result of replacing two occurrences of $17n$ with $m.$ The 2nd equality is the result of replacing $n$ with $m/17$ (note that from $m = 17n$ we get $m/17 = n).$ The 3rd equality is by arithmetic (division can be accomplished by inverting and multiplying).
At this point we have the limit you wanted, except that the letter $m$ is used instead of the letter $n,$ and $17$ is being used instead of $x.$ However, since this letter is a dummy variable in the present context (see also this MSE answer), we have
$$ \lim_{m \rightarrow \infty} \left(1 + \frac{x}{m} \right)^{m} \; = \;\; \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n} \right)^{n} \; = \;\; \lim_{u \rightarrow \infty} \left(1 + \frac{x}{u} \right)^{u} \; = \;\; \lim_{@ \rightarrow \infty} \left(1 + \frac{x}{@} \right)^{@} \; = \;\; \cdots $$
Of course, in the above we wouldn't want to use the letter $x$ for the dummy variable. Knowing which symbols we CAN'T replace a dummy variable by in a given expression is usually obvious in practice, but if you're interested, the precise restrictions involve the notion of free and bound variables.