Limit x·LN(x) as x→0⁺

**MathCrusader** · January 20th 2013, 01:47 AM

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.

**a tutor** · January 20th 2013, 03:02 AM

$x \ln x = \frac{\ln x}{1/x}$ .

**MathCrusader** · January 20th 2013, 04:47 AM

Originally Posted by a tutor

$x \ln x = \frac{\ln x}{1/x}$ .

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

**zzephod** · January 20th 2013, 05:25 AM

Originally Posted by MathCrusader

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:

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

.

**MathCrusader** · January 20th 2013, 11:34 AM

Originally Posted by zzephod

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

Hence:

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

.

Ah, I stand corrected. I was using the rule incorrectly; instead of differentiating the divisor och dividend respectively, I instead used the quotient rule! Thank you!

Thread: Limit x·LN(x) as x→0⁺

LinkBack

Thread Tools

Display

Limit x·LN(x) as x→0⁺

Re: Limit x·LN(x) as x→0⁺

Re: Limit x·LN(x) as x→0⁺

Re: Limit x·LN(x) as x→0⁺

Re: Limit x·LN(x) as x→0⁺

Similar Math Help Forum Discussions

f: E → (X1 x X2) is cont iff (g1 o f): E → X1 and (g2 o f): E → X2 are both cont

why does lim(z→e^(πi/4)) (z - e^(πi/4))/(1 + z^4) =...

Analysis help--> Suppose that g1: A1 → A2 and g2 : A2 → A3. If B ⊂ A3...

is f: x → |x| a function of x?

proof of a limit as n→∞

Search Tags