Prove $e=\lim\limits_{x \to \infty}(1+\frac1x)^x$ 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.
If you have
$\log(1+x)
=\int_0^x \dfrac{dt}{1+t}
$,
then,
if $x &gt; 0$,
$\log(1+x)
=\int_0^x \dfrac{dt}{1+t}
\lt\int_0^x dt
=x$
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.
If you have
$\log(1+x)
=\int_0^x \dfrac{dt}{1+t}
$,
then,
if $x &gt; 0$,
$\log(1+x)
=\int_0^x \dfrac{dt}{1+t}
\lt\int_0^x dt
=x$
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.