Calculate
$$\lim_{x\to 0}\frac{e^{\frac{\ln{(1+x)}}{x}}-e}{x}$$
My Attempt:
$$\lim_{x\to 0}e\cdot \frac{e^{\frac{\ln{(1+x)}}{x}-1}-1}{\frac{\ln{(1+x)}}{x}-1} \cdot \frac{\frac{\ln{(1+x)}}{x}-1}{x}$$
$$\lim_{x\to 0}e.\frac{\frac{\ln{(1+x)}}{x}-1}{x}$$
I am not able to solve this further. I can easily use L-Hospital to show $$\lim_{x\to 0}\frac{\frac{\ln{(1+x)}}{x}-1}{x}=-\frac{1}{2}$$
How can I prove this without L-Hopital or Taylor expansions?
$$\ln(1+x)=x-\dfrac12x^2+\dfrac13x^3+\cdots$$ then \begin{align} e^{\frac{\ln(1+x)}{x}} &= e^{1-\frac12x+\frac13x^2+\cdots} \\ &= e\left(e^{-\frac12x+\frac13x^2+\cdots}\right) \\ &= e\left(1+\left(-\frac12x+\frac13x^2+\cdots\right)+\dfrac12\left(-\frac12x+\frac13x^2+\cdots\right)^2+\cdots\right) \\ &= e\left(1-\dfrac12x+O(x^2)\right) \end{align} then $$\lim_{x\to 0}\frac{e^{\frac{\ln{(1+x)}}{x}}-e}{x}=\lim_{x\to 0}\frac{e\left(1-\dfrac12x+O(x^2)\right)-e}{x}=\color{blue}{-\dfrac12e}$$