Why is $\lim\limits_{x \to 0^+}(1+ x)^\frac1x$ is $e$ and not $\infty$?

Solution 1:

The issue is that the first statement should really be

If $a$ is a constant greater than $1$, then $\lim_{x\to\infty}a^x=\infty$.

It is normally understood that if you say "Let $a>1$", or similar, that you are talking about a specific, constant real number $a$.

Here, it's important that $a$ does not depend on $x$, and without that restriction the statement wouldn't be true. It's easier to see this by replacing $a$ with $\sqrt[x]2$; then certainly $\sqrt[x]2>1$ for any $x$, but $(\sqrt[x]2)^x=2$ for every $x$, so its limit is also $2$.

Solution 2:

$$ \left(1+\frac12\right)^2=1+\frac22+\frac14<\frac1{0!}+\frac1{1!}+\frac1{2!}\\ \left(1+\frac13\right)^3=1+\frac33+\frac39+\frac1{27}<\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}\\ \left(1+\frac14\right)^4=1+\frac44+\frac6{16}+\frac4{64}+\frac1{256}<\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}\\ \left(1+\frac15\right)^5=1+\frac55+\frac{10}{25}+\frac{10}{125}+\frac5{625}+\frac1{3125}<\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}\\ $$

as you can check term-wise, and this generalizes to any power. The series on the right is quickly convergent to $2.718281828\cdots$