Re: Limit x·LN(x) as x→0⁺

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

Re: Limit x·LN(x) as x→0⁺

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Originally Posted by

**a tutor** $\displaystyle x \ln x = \frac{\ln x}{1/x}$.

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

Re: Limit x·LN(x) as x→0⁺

Quote:

Originally Posted by

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

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

Hence:

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

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Re: Limit x·LN(x) as x→0⁺

Quote:

Originally Posted by

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

Hence:

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

.

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!