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Math Help - Limit xLN(x) as x→0⁺

  1. #1
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    Limit xLN(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'Hspitals 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.
    Last edited by MathCrusader; January 20th 2013 at 01:50 AM.
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  2. #2
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    Re: Limit xLN(x) as x→0⁺

    x \ln x = \frac{\ln x}{1/x}.
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    Re: Limit xLN(x) as x→0⁺

    Quote Originally Posted by a tutor View Post
    x \ln x = \frac{\ln x}{1/x}.
    Beknownst to me, indeed. However, the result remains the same.
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    Re: Limit xLN(x) as x→0⁺

    Quote Originally Posted by MathCrusader View Post
    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}}= ?

    .
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  5. #5
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    Re: Limit xLN(x) as x→0⁺

    Quote Originally Posted by zzephod View Post
    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!
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