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?
On
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 $e.$
You need the formal definition for limit to explain this exactly, but here's an intuitive way to see it.
The expression $$ \left( 1 + \frac{1}{n} \right)^{n} $$ whose limit is $e$ is subtle. The sum inside the parentheses approaches $1$, which tends to make the limit $1$. But the exponent grows to infinity. So you are raising numbers nearer and nearer to $1$ to higher and higher powers. The result might be anything between $1$ and infinity, depending on how those conflicting tendencies fight it out. In this case the compromise turns out to be $e \approx 2.718$.
Now on to your question. Very loosely speaking, $2n$ grows to infinity twice as fast as $n$ does. So the exponent in $$ \left( 1 + \frac{1}{n} \right)^{2n} $$ is making the limit bigger faster than the decrease in $1/n$ is making it smaller. To see why the actual limit is $e^2$ simply rewrite $$ \left( 1 + \frac{1}{n} \right)^{2n}= \left[ \left( 1 + \frac{1}{n} \right)^{n} \right]^2 . $$