If $e$ (Euler's constant) $= (1+\frac{1}{n})^n$ as $n$ approaches infinity, why is $e^x$ not equal to $e$? if x equals any number, real or not real. If $e$ (Euler's constant) $= (1+\frac{1}{n})^n$ as $n$ approaches infinity, why is $e^x$ not equal to $e$ if x equals any number, real or not real.
Let r = any real number
I think that $\left(1+\frac{1}{n}\right)^{rn} = e^r$, and $rn = n$ when $n$ approaches infinity, so $e^r$ should equal $e$.
Also, $0*infinity=1$, so if both sides are raised to the $e$, you get 1$*$e^infinity=$e$, so $e$^infinity should also equal $e$.
Could anyone tell me what I did wrong here?

Just as $\lim \frac{2x}{x} \neq \lim \frac{x}{x},$ despite both fractions being formally $\infty/\infty.$ It’s not enough to notice that numerator and denominator are both $\infty$. You need to know they go to infinity at the same rate, to conclude that the ratio is one. 
Similarly in the limit $\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{2n},$ if your exponent approaches infinity twice as fast as your base’s discrepancy from $1$ approaches zero, that must be accounted for. Those rates must be exactly the same for the resulting limit to be exactly $e.$
