Proving that $\lim_\limits{x \to \infty} x\cdot \ln\left(1 + \frac{1}{x}\right) = 1$ using Taylor series We need to prove that $\lim_\limits{x \to \infty} \ln((1 + \frac{1}{x})^x) = \lim_\limits{x \to \infty} x \cdot \ln(1 + \frac{1}{x}) = 1$ and would like to use Taylor series.
To do this, when we expand $\ln(1 + \frac{1}{x})$, we should get a series like $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$, multiplying by $x$ would make it $1 + \frac{a_1}{x} + \frac{a_2}{x^2} + ...$ which is $1$ if $x \to \infty$.
The Taylor expansion of $f(x) = \ln(1 + \frac{1}{x})$ at some $a$ is $\sum_{n = 0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{6}(x-a)^3 + ...$.
$f'(a) = -\frac{1}{a^2 + a}$, $f'(a) \cdot (x-a) = - \frac{x-a}{a^2 + a}$
$f''(a) = \frac{2a + 1}{(a^2 + a)^2}$, $f''(a) \cdot (x-a) = \frac{(2a + 1)(x-a)}{(a^2 + a)^2}$
So the Taylor series begins like this:
$\ln(1 + \frac{1}{x}) = \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ...$
Since the first term doesn't involve $x$ and we want it to be $\frac{1}{x}$, we would like it to be $0$ and the next term to be $\frac{1}{x}$. To make the first term $0$, $a$ needs to be approaching infinity.
$\lim_\limits{a \to \infty} \left( \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right) = \lim_\limits{a \to \infty} \left(- \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right)$.
Now, this isn't in the form $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$ that we wanted.
Do we need to pick some other $a$?
I found https://math.stackexchange.com/a/1071689/1095885 which says

When x is very large, using Taylor
$\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\left(\frac{1}{x}\right)^4\right)$

The series $\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{x^3} + ...$, which would be perfect, seems to be a Laurent series though.
Edit: When substituting $u = \frac{1}{x}$ and taking the Taylor series of $\ln(1 + u)$, why don't we have to use the chain rule?
 A: It seems that you're making this a bit more complicated than is necessary.
For $\lvert u \rvert < 1$,
$$
\ln(1 + u) = u - \frac12 u^2 + \frac13 u^3 - \frac14 u^4
+ \cdots 
$$
You can verify this Maclaurin series (Taylor series centered at $0$) by differentiating both sides to obtain the well-known geometric series
$$
\frac{1}{1+u} = 1 - u + u^2 - u^3 + \cdots  
$$
and by verifying that the constant term vanishes by evaluating at $u=0$.
Now, by dividing through by $u$, we have the series
$$
\frac1u \ln(1 + u) = 1 - \frac12 u + \frac13 u^2 - \frac14 u^3
+ \cdots 
$$
which has the same radius of convergence, using, say the Alternating Series Test.
As $x \to \infty$, we can certainly bound $\lvert x \rvert > 1$, so that
$\lvert u \rvert = \lvert \frac1x \rvert < 1$, hence
$$
x \cdot \ln\biggl(1 + \frac1x\biggr) 
= 1 - \frac{1}{2x} + \frac{1}{3x^2} - \frac{1}{4x^3}
+ \cdots 
$$
converges too for any such $x$.
It's easy to see that all but the constant term vanish as $x \to \infty$, revealing the limit.
A: Using Taylor series here is like the proverbial killing a fly with a sledgehammer. Set $h = 1/x$, so the condition $x \to \infty$ is the same as $h \to 0^+$ and
$$
x\ln\left(1+\frac{1}{x}\right) = \frac{\ln(1+h)}{h} = \frac{\ln(1+h)-\ln 1}{h}.
$$
Therefore
$$
\lim_{x \to \infty} x\ln\left(1+\frac{1}{x}\right)  = 
\lim_{h \to 0^+} \frac{\ln(1+h)-\ln 1}{h}, 
$$
which shows your calculation follows from calculating the derivative of $\ln x$ at $x = 1$ (using $h \to 0$, not just $h \to 0^+$):
$$
\lim_{h \to 0} \frac{\ln(1+h)-\ln 1}{h} = \left.\frac{d}{dx}\right|_{x=1}\ln x = \ ??
$$
A: An elementary way to do
the discrete version of this
is to show that
if
$a_n=(1+1/n)^n$
and
$b_n=(1+1/n)^{n+1}$
then
$a_n$ is increasing and
$b_n$ is decreasing
and,
as $n \to \infty$,
$a_n$ and $b_n$
have a common limit,
which
I suggest
we call $e$.
An elementary proof of this,
which only uses the
arithmetic mean-geometric mean inequality,
can be found in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58
(1951), 563.
