calculus - Evaluating $lim_{x to 0^+} (e^x-1)^{frac{(tan{x})^2}{sqrt[3]{x^2}}}$

I cannot figure this limit out.

$$\lim_{x \to 0^+} (e^x-1)^{\frac{(\tan{x})^2}{\sqrt[3]{x^2}}}$$

I've used the e to the ln trick and multiplied by 1 ($\frac{x^2}{x^2}$) and arrived at

$$\lim_{x \to 0^+} \exp({x^{4/3}} \ln ({e^x-1})) $$

However I failed at getting further. I tried adding and subtracting $\ln x$ but that got me nowhere.

I cannot use l'Hospital or Taylor series (only the "known" limits for $\sin$, $\cos$, $e^x$, $\ln$ such as $\lim_{x \to 0}\frac{sinx}{x}=1$ which are really only Taylor series).

Thanks for help!

Answer

Hint: $$\lim_{x\to 0}\frac{e^x-1}x=\lim_{x\to 0}\frac{\tan x}x=1$$