$h(x)$ is a continuous function in the section $I$, $0 \in I$, $h(x)$ has a derivative at $x=0$ and $h(0)=0, h(x)>-1 (0 \ne x \in I)$. $f(x) = (\frac{(1+h(x))^{1/h(x)}}{e})^{1/x}$, for every $0 \ne x \in I$. I need to prove that $$ \lim_{x\to0}f(x)= \sqrt{\frac{1}{e^{h'(0)}}}=e^{-\frac{h'(0)}{2}}$$.
$$\lim_{x\to 0}{f(x)} = \lim_{x\to 0}{e^{ln{f(x)}}}$$ hence I need to find $\lim_{x \to 0}{ln f(x)}$. $$\lim_{x \to 0}{ln f(x)} = \lim_{x \to 0}{ln (\frac{(1+h(x))^{1/h(x)}}{e})^{1/x}}=\lim_{x \to 0}{\frac{ln(\frac{(1+h(x))^{1/h(x)}}{e})}{x}}=\lim_{x \to 0}{\frac{ln((1+h(x))^{1/h(x)}-ln (e)}{x}}=\lim_{x \to 0}{\frac{\frac{1}{h(x)}ln(1+h(x))-1}{x}}$$ I then expressed $h(x)$ using Maclaurin series, with Lagrange form of the remainder, to express $h(x)$ and $ln (1+h(x))$ I get: $$h(x)=h'(\xi)\cdot x$$ and $$ln (1+h(x))=\frac{h'(\xi)}{1+h(x)}=\frac{h'(\xi)\cdot x}{1+h'(\xi)\cdot x}$$ and getting back to the limit $$\lim_{x \to 0}{\frac{\frac{1}{h'(\xi)\cdot x}\frac{h'(\xi)\cdot x}{1+h'(\xi)\cdot x})-1}{x}}=\lim_{x \to 0}{\frac{-h'(\xi)\cdot x}{x+x^2\cdot h'(\xi))}}$$ and using l'hopital's rule I get $$\lim_{x \to 0}{\frac{-h'(\xi)\cdot x}{x+x^2\cdot h'(\xi))}}=\lim_{x \to 0}{\frac{-h'(\xi)}{1+2x\cdot h'(\xi))}}$$ because $\xi\to 0$ when $x\to0$ the \limit I finally get is $$\lim_{x \to 0}{\frac{-h'(\xi)}{1+2x\cdot h'(\xi)}} =\frac{-h'(0)}{1+2\cdot 0 \cdot h'(0))}=-h'(0)\Longrightarrow \lim_{x\to0}{f(x)}=\frac{1}{e^{h'(0)}}$$ I tried working around this so many times but I don't know how I get the root part of the wanted result.
EDIT
I tried a different approach $$\lim_{x \to 0}{\frac{ln(1+h(x))-h(x)}{x\cdot h(x)}}$$
using the Maclaurin series: $$h(x) = h(0) + h'(0)x + \frac{h''(0)x^2}{2} + R_2$$ $$h(x) = h(0) + h'(0)x + R_1$$ $$ln (1 + h(x)) = h'(0)x + \frac{h''(0)x^2 - (h'(0))^2 x^2}{2} + S_2$$ ($R_2$, $S_2$ are the remainders of order 2 and $R_1$ is the remainder of order 1). then $$\lim_{x \to 0}{\frac{ln(1+h(x))-h(x)}{x\cdot h(x)}}=\lim_{x \to 0}{\frac{h'(0)x + \frac{h''(0)x^2 - (h'(0))^2 x^2}{2} + S_2-h(0) - h'(0)x - \frac{h''(0)x^2}{2} - R_2}{x(h(0) + h'(0)x + R_1)}}=\lim_{x \to 0}{\frac{-\frac{(h'(0))^2 x^2}{2}+S_2 - R_2}{x^2h'(0)+xR_1}}=\lim_{x \to 0}{\frac{-\frac{(h'(0))^2}{2}+\frac{S_2}{x^2} - \frac{R_2}{x^2}}{h'(0)+\frac{R_1}{x}}}$$ now because $$\lim_{x \to 0}{\frac{S_2}{x^2}}=0, \lim_{x \to 0}{\frac{R_2}{x^2}}=0, \lim_{x \to 0}{\frac{R_1}{x}}=0$$ the limit is $$\lim_{x \to 0}{\frac{-\frac{(h'(0))^2}{2}+\frac{S_2}{x^2} - \frac{R_2}{x^2}}{h'(0)+\frac{R_1}{x}}}=\frac{-\frac{(h'(0))^2}{2}}{h'(0)}=-\frac{h'(0)}{2}$$ hence $$ \lim_{x\to0}f(x)= \sqrt{\frac{1}{e^{h'(0)}}}=e^{-\frac{h'(0)}{2}}$$ but I don't know if I'm allowed to use 2 different order of Maclaurin series
This can come straight from the definition of the derivative and the limit: Since $h(0) = 0$ and letting $k = h'(0)$, we have that for every $\epsilon > 0$ there is a $\delta > 0$ such that if $|x| < \delta$, then $\left|\frac{h(x) - 0}x - k\right| < \epsilon$, so $$kx - \epsilon|x| < h(x) < kx + \epsilon|x|$$
That is all you need to know to take your limit.