The subject is Taylor and Malcaurin polynomials.
Let $h(x)$ be a continuous function on the interval $I\ni x=0$.
Differentiable at $x = 0$
And: $$h(x):=\begin{cases}0,&x=0\\y\in(-1,+\infty)\setminus\{0\},&x\in I\setminus\{0\}\end{cases} $$
Define:
$$ f(I\setminus\{0\}) = \left(\frac{(1+h(x))^{\frac{1}{h(x)}}}{e}\right)^{\frac{1}{x}}$$ Show that: $$ \lim_{x \to 0}f(x) =\sqrt{\frac{1}{e^{h'(0)}}} $$
The hint for the ex. was to take $\ln(f(x))$
What I did:
As the hint suggested, I did $\ln(f(x))$
And got to: $$ \ln(f(x)) = \frac{1}{xh(x)}\ln(1+h(x)) - \frac{1}{x} = \frac{1}{x}\left(\frac{1}{h(x)}\ln(1+h(x)) - 1\right) $$
I looked at the term I need to get to. I see $h'(0)$.
Therefore, I developed a Maclaurin polynomial:
$$ P_n(x) = h(0) + h'(0)x + R_n(x) = 0 + h'(0) + R_n(x) $$
And now I am pretty stuck.
I thought to make a Maclaurin for $\ln(1+h(x))$ instead, maybe, but also, I don't see it bringing me anywhere.
Can I have a hint?
Thank you in advance!
Starting from $$\ln(f(x))=\frac{1}{xh(x)}(\ln(1+h(x))-h(x))$$ and using the Taylor series at $x=0$ for $\ln(1+x)$, that is, $$\ln(1+x)=\sum_{n\geq 0}\frac{(-1)^nx^{n+1}}{n+1}$$ you have that $$\frac{1}{xh(x)}(\ln(1+h(x))-h(x))=\frac{1}{x}\left(-\frac{h(x)}{2}+h(x)F(h(x))\right)=-\frac{h(x)}{x}+\frac{h(x)}{x}F(h(x))$$ where $F(h(x))\to 0$ when $x\to 0.$ Because $h$ is differentiable at $x=0$, and $h(0)=0$ you get $$\lim_{x\to 0}\ln f(x)=-\frac{h'(0)}{2}.$$