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

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.

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

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?

• What are you allowed to use? Your 'rewrite' doesn't actually work since $0\ln(0)$ isn't defined (and you haven't showd that $\lim_{x\to0}x\ln x=0$...) – Steven Stadnicki Mar 7 '15 at 18:02
• It might help you to observe that $\frac{e^{x\ln(x)}-1}{x}=\frac{(e^{\ln(x)})^x-1}{x}=\frac{x^x-1}{x}$... – barak manos Mar 7 '15 at 18:04
• while $lim_{x \rightarrow 0} f(x)$ isn't defined, $lim_{x \rightarrow 0^+} f(x)$ is still ok. – Blex Mar 7 '15 at 18:06
• @Educ It gives all the details required - if anything, I should have left out some details, but I didn't. What bit don't you understand? – Sam T Mar 7 '15 at 18:51
• Just an observation but I've wanted to get this off my chest for a while: what exactly is achieved by questions proscribing the use of this technique or that? I can sort of see the utility if one is trying to 'block' a quick solution to a cleverly-set elementary Euclidean geometry problem using trigonometry, for instance, but the pedagogical value in forbidding LHR (and even series expansions here!) in limit problems totally escapes me. If they wanted a solution by using the limit definition of a derivative, they should just say that explicitly. – Deepak Mar 8 '15 at 11:47

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 $x=0$.

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! :)

• "However, taking the derivative for x>0, then taking the limit should work." Ouch! The limit at 0 of the derivative may not exist while the derivative at 0 does. – Did Mar 8 '15 at 11:18
• Ah, fair point, not great wording. Updated! :) - thanks for your comment! – Sam T Mar 8 '15 at 11:30
• Does defining $g(x) = \int_{-\infty}^x G(t) dt$ with $G(t) = 0$ for $t \le 0$ and $G(t) = 1/t$ for $t>0$ provide a counterexample... can't just apply the FTC since $G$ isn't continuous... - if not, what would? :) – Sam T Mar 8 '15 at 11:36

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 $h<<1$

• $e^{x\ln(x)}=x^{x}$ – Educ Mar 8 '15 at 15:07