# How to calculate this limit without L'Hopital rule?

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.

• Multiply and divide by $\log x$ Nov 16 '18 at 14:40
• e^(xln(x)) = x^x might help Nov 16 '18 at 20:55

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

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

HINT

The key point is that $$x\log x \to 0$$ then

$$\frac{e^{x\ln(x)}-1}{x}=\frac{e^{x\ln(x)}-1}{x\ln(x)}\ln x$$