Calculating limit using MacLaurin series I'm asked to calculate the following limit using Maclaurin series.
$$\lim_{x\to 0}\frac{\left(1 + x\right)^{1/x} - e} { x}$$
They say that a good piece of advice is this: $f(x)^g(x) = e^{g(x)  \log(f(x)}$
My work:
$$\begin{align}
\frac{\left(1 + x\right)^{1/x} - e} { x}&=\frac{e^{\frac1x\log(1+x)} - e} { e^{\log(x)} }\\\\
&=\frac{1 + \frac1x\log(1+x) - e} { 1 + \log(x)}
\end{align}$$
??? Now what? I have been struggling with this exercise for two days, haha. (Working full time though)
 A: Well, writing $\left(1+x\right)^{1/x}=e^{\frac1x\log(1+x)}$ was fine.  But writing $x=e^{\log(x)}$, while correct, doesn't facilitate the analysis.
So, let's use the Taylor Series $\log(1+x)=x-\frac12x^2+O(x^3)$ and $e^x=1+x+O(x^2)$ and write
$$\begin{align}
\frac{\left(\left(1+x\right)^{1/x}-e\right)}{x}&=\frac{e^{\frac1x\log\left(1+x\right)}-e}{x}\\\\
&=\frac{e^{\frac1x\left(x-\frac12x^2+O(x^3)\right)}-e}{x}\\\\
&=e\,\left(\frac{e^{-\frac12x+O(x^2)}-1}{x}\right)\\\\
&=e\,\left(\frac{-\frac12x+O(x^2)}{x}\right)\\\\
\end{align}$$
Can you finish now?
A: The Taylor development will turn the denominator to a polynomial (plus a remainder). There is no reason to transform the denominator, which is already a polynomial.
Now,
$$(1+x)^{1/x}=e^{\log(1+x)/x}=e^{(x-x^2/2+x^3/3\cdots)/x}=e^{1-x/2+x^2/3\cdots}=e\cdot e^{-x/2+x^2/3\cdots}
\\=e\left(1+\left(-\frac x2+\frac{x^2}3-\cdots\right)+\frac12\left(-\frac x2+\frac{x^2}3-\cdots\right)^2+\cdots\right)$$
and
$$\frac{(1+x)^{1/x}-e}x=e\frac{-\dfrac x2+O(x^2)}x.$$
