Prove $e=\lim\limits_{x \to \infty}(1+\frac1x)^x$

Question

The same exercise requires us to prove that $\lim_{x \to \infty}x\log(1+\frac1x)=1$ just before proving the identity in the title. I can use this fact as follows:

$$\lim_{x \to \infty}\left(1+\frac1x\right)^x=\lim_{x \to \infty}e^{x\log(1+\frac1x)}=e$$

Is this proof valid? I'm asking because the author is using a complicated $\varepsilon-\delta$ proof which could have been avoided, if I'm not mistaken.

This depends on what you are allowed to use. Yet, this proof would be fine. — VIVID, Apr 04 '21 at 17:16
They have not disallowed to use the identity $a^x=e^{x\log{a}}$ — super.t, Apr 04 '21 at 17:17
You do need to show that you can move the limit to the exponent. That requires showing that exponentiation is continuous. But yeah, your proof is good. — Eric, Apr 04 '21 at 17:46
@Eric now I see that it was the point of the $\delta-\varepsilon$ proof. — super.t, Apr 04 '21 at 18:12
Here's my fave https://math.stackexchange.com/a/4074292/403337 — , Apr 04 '21 at 18:31

score -1 · Answer 1 · answered Apr 04 '21 at 18:25

If you have $\log(1+x) =\int_0^x \dfrac{dt}{1+t} $, then, if $x > 0$, $\log(1+x) =\int_0^x \dfrac{dt}{1+t} \lt\int_0^x dt =x$ and $\log(1+x) \gt \int_0^x \dfrac{dt}{1+x} = \dfrac{x}{1+x} = \dfrac{x+x^2-x^2}{1+x} = x-\dfrac{x^2}{1+x} $ so $0 \gt \dfrac{\log(1+x)}{x}-1 \gt -\dfrac{x}{1+x} $ so $\lim_{x\to 0}\dfrac{\log(1+x)}{x} =1 $.

This is probably your proof.

Prove $e=\lim\limits_{x \to \infty}(1+\frac1x)^x$

1 Answers1