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$$\displaystyle\lim_{x\to 0^{\hspace{.02 in}+}} \: (x\ln(x)) =\ ?$$

This should be $0$ times $-\infty$, so I believe it's indeterminate in this form, but I don't know how to solve the problem any further than this, if it is possible.

Thank you for any help.

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It's zero.

Trivial explanation: the logarithm is a very sloooow function, whereas $x$ is not, hence $x$ dominates.

Non trivial explanation: you can use Hôpital rule after having rewritten it in a smart way as

$$\lim_{x\to 0^+} \frac{\ln(x)}{\frac{1}{x}}$$

A brutal substitution leads you to a Hôpital form $-\infty/\infty$. Applying Hôpital:

$$\lim_{x\to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x\to 0^+} -x = 0$$

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You can rewrite this as

$$\lim_{x\to0^+}\frac{\ln x}{\frac1x}$$ and use L'Hospital.

You get

$$\lim_{x\to0^+}\frac{\frac1x}{-\frac1{x^2}}$$

which should be trivial to calculate.

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L'Hospital's rule is not the swiss knife of limits computations!

The simplest is to make the $x=\dfrac1t$ ($t\to+\infty$) substitution: $$x\ln x=-\frac{\ln t}t\xrightarrow[t\to+\infty]{}0,$$ the latter being a standard high school limit.

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  • $\begingroup$ How did you go from x*ln(x) to ln(x)/x in your 2nd equation? $\endgroup$ – Aaron Franke Feb 1 '17 at 10:39
  • $\begingroup$ Properties of log: $\ln x=\ln\dfrac1t=-\ln t$, that's all. $\endgroup$ – Bernard Feb 1 '17 at 10:43
  • $\begingroup$ What does the t variable stand for? $\endgroup$ – Aaron Franke Feb 1 '17 at 21:15

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