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$$\lim _{x \rightarrow 0} x \ln(x)$$

I would say the limit will equal 1+ ln(x) but x cannot equal 0 at this point, so what shall i do ?

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  • $\begingroup$ Could you say more about why you think the limit equals 1 + ln(x) (and what you even mean by that)? $\endgroup$ – mweiss Jul 8 '14 at 2:02
  • $\begingroup$ $1+\ln x$ is the derivative of $x\ln x$, but l'Hopital's rule for $\lim f(x)$ has nothing to do with the derivative of $f(x)$. $\endgroup$ – David Jul 8 '14 at 2:12
  • $\begingroup$ Remember that L'Hopital rule applies when you evaluate a limit and the result is an undetermined form ($0/0, \infty/\infty$) $\endgroup$ – AndreGSalazar Jul 8 '14 at 4:16
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You need to turn it into something that you can use l'Hopital on - that is, 0/0 or $\infty/\infty$.

For example, $$\frac{\ln(x)}{1/x}$$

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  • $\begingroup$ So does l'Hopital's rule apply? $\endgroup$ – John Jul 8 '14 at 1:59
  • $\begingroup$ Well, what are the limits of the numerator and denominator of that expression as x approaches 0 from the right? $\endgroup$ – mweiss Jul 8 '14 at 2:01
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Let $u=-\ln(x)$ which changes the problem to

$$- \lim_{u\to \infty} \frac{u}{e^u}.$$

Can you see it now?

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$$\lim _{x \rightarrow 0} x \ln(x)=\lim _{x \rightarrow 0} \frac {\ln(x)}{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\frac{1}{x}}{\frac{-1}{x^2}}=-\lim _{x \rightarrow 0}x=0 $$

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without L'Hopital

$\lim _{x \rightarrow 0} x \ln(x) = \lim_{x \rightarrow 0} \frac{\ln(x)}{1/x}$

then, $ y = 1/x $ and when $x \rightarrow 0$, we know $y \rightarrow \infty$

then our limit is $\lim_{y \rightarrow \infty} \frac{\ln(1/y)}{y} = \lim_{y \rightarrow \infty} \frac{\ln(1)- \ln(y)}{y} = \lim_{y \rightarrow \infty} \frac{- \ln(y)}{y} = \lim_{y \rightarrow \infty} -\frac{1}{y} \ln(y)$

$\lim_{y \rightarrow \infty} -\ln(y ^ \frac{1}{y} )$

And We know $\lim_{y \rightarrow \infty} y ^ \frac{1}{y} = 1 $ , then our limit
$\lim_{y \rightarrow \infty} -\ln(y ^ \frac{1}{y} ) = -\ln(1) = 0$

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