The limit $\lim_{x\to 0}\frac{e^{x\ln(x)}-1}{x}$

Question

I would like to show that

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

without using:

• L'Hôpital's rule
• expansion series.

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My thoughts

$$ \lim_{x\to 0^{+}}\frac{e^{x\ln(x)}-1}{x}=\lim_{x\to 0^{+}}\frac{\ln(x)(e^{x\ln(x)}-1)}{x\ln(x)}=\left(\lim_{x\to 0^{+}}\ln(x)\right)\left(\lim_{x\to 0^{+}}\frac{e^{x\ln(x)}-1}{x\ln(x)}\right)$$

• $\left(\lim_{x\to 0^{+}}\ln(x)\right)=-\infty$
• $\left(\lim_{x\to 0^{+}}\frac{e^{x\ln(x)}-1}{x\ln(x)}\right)$

let $x\ln(x)=t$ then we have

$$\left(\lim_{x\to 0^{+}}\frac{e^{x\ln(x)}-1}{x\ln(x)}\right)=\left(\lim_{t\to 0^{+}}\frac{e^{t}-1}{t}\right)=1$$

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

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Questions:

Am I right?
Is there any other way that let us to calculate that limit without using L'Hôpital's rule or any expansion series?

Answer 1

As you've shown, it's the derivative at $0$*. So I would recommend just differentiating $\exp(x\ln(x))$, seeing what you get and then taking the limit $x \to 0$. $$\frac{d}{dx} [\exp(x \ln(x))] = (\ln(x) + 1)\exp(x\ln(x)) \to -\infty, \ x \to 0^+,$$ since $x\ln(x) \to 0$ as $x \to 0^+$, and $\exp$ is continuous, so $\exp(x\ln(x)) \to 1$ as $x \to 0^+$.

*It isn't actually the derivative at $0$, since $\ln(0)$ is not defined. However, you are looking for the limit as $x \to 0$ (since $\ln(0)$ is not defined), so you take the limit $x \to 0^+$.

Hope this helps! :)

Answer 2

The last limit you wrote is the definition of the derivative of the function $f(x) = \ e^{x\;lnx}$ for $x=0$ . Use the chain rule to obtain the derivative $f´(x)$ and put $x=0$.

Answer 3

Hint: $$e^{ln(a)}=a$$

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

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

$$=\ - \infty$$

here $0^0$ won't occur, as $x=0+h$, where $h<<1$