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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.

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

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

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HINT

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

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

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