Find $\lim_{x \to 0^{+}} \frac{\pi^{x\ln x} - 1}{x}$ if it exists . Let $f(x) = \frac{\pi^{x\ln x} - 1}{x}$ . Find $\lim_{x \to 0^{+}}f(x)$ if it exists . 
My try : $f(x) = \frac{\pi^{x\ln x} - 1}{x} = \frac{e^{x\ln x \ln \pi} - 1}{x}$ . Using $(\forall u\in\mathbb{R}):e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots$ and putting $u = {x\ln x \ln \pi} $ leads to $\lim_{x \to 0^{+}}f(x) = -\infty$ . I'm not sure whether or not my answer is right . Also I'm looking for other solutions .

I think your answer is right.
Since $x\ln{x}\rightarrow0$, we obtain:$$\frac{\pi^{x\ln{x}}-1}{x}=\frac{e^{x\ln{x}\ln\pi}-1}{x\ln{x}\ln\pi}\cdot\ln\pi\ln{x}\rightarrow-\infty$$
