Calculate this limit without L’Hopital rule

Question

I want to evaluate the following limit without using the L'Hopital rule : $$ \lim\limits_{x\rightarrow 0^+}\frac{e^{x\ln(x)}-1}{x}$$
I know the answer is $-\infty$.
I can demonstrate that graphically and by using the L'Hopital rule.
Any hint would be appreciated and thanks.

Answer 1

Note that for $x\log(x)<1$,

$$e^{x\log(x)}\le \frac1{1-x\log(x)}$$

whereby we see that for $x\le 1$

$$\begin{align} \frac{e^{x\log(x)}-1}{x}&\le \frac{\log(x)}{1-x\log(x)}\\\\ &\le \frac{e}{e+1}\,\log(x) \end{align}$$

Inasmuch as $\log(x)\to -\infty$, we find that

$$\lim_{x\to 0^+}\frac{e^{x\log(x)}-1}{x}=-\infty$$