# Limit x·LN(x) as x→0⁺

• Jan 20th 2013, 02:47 AM
Limit x·LN(x) as x→0⁺
So, we're faced with this:

$\lim_{x \rightarrow 0^+} \, x\ln (x) \, .$

Since we get an indeterminate form when attempting to evaluate at $x = 0$, we resort to L'Hôspitals rule. The derivative of $x\ln (x)$ is $\ln (x) + 1$, therefore

$\lim_{x \rightarrow 0^+} \, x\ln (x) = \lim_{x \rightarrow 0^+} \, \ln (x) + 1 \, .$

I am stuck here, as it eludes me how this approaches 0? I get negative infinity, but apparently that's incorrect.
• Jan 20th 2013, 04:02 AM
a tutor
Re: Limit x·LN(x) as x→0⁺
$x \ln x = \frac{\ln x}{1/x}$.
• Jan 20th 2013, 05:47 AM
Re: Limit x·LN(x) as x→0⁺
Quote:

Originally Posted by a tutor
$x \ln x = \frac{\ln x}{1/x}$.

Beknownst to me, indeed. However, the result remains the same.
• Jan 20th 2013, 06:25 AM
zzephod
Re: Limit x·LN(x) as x→0⁺
Quote:

Beknownst to me, indeed. However, the result remains the same.

No it does not, $\frac{d}{dx}\ln(x)=\frac{1}{x}$ while $\frac{d}{dx}\frac{1}{x}=-\;\frac{1}{x^2}$

Hence:

$\lim_{x \to 0^+}\frac{ \frac{d}{dx}\ln(x)}{\frac{d}{dx}\frac{1}{x}}= ?$

.
• Jan 20th 2013, 12:34 PM
No it does not, $\frac{d}{dx}\ln(x)=\frac{1}{x}$ while $\frac{d}{dx}\frac{1}{x}=-\;\frac{1}{x^2}$
$\lim_{x \to 0^+}\frac{ \frac{d}{dx}\ln(x)}{\frac{d}{dx}\frac{1}{x}}= ?$