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Prove that $\lim_{x\to0^{+}} (-x) \ln x$ exists and find it.

I know we should write $(-x)*\ln(x)$ as $\ln(x)/1/x$. Then we can see by L'Hopital's Rule that it equals $x$. And as $x$ approaces $0^{+}$ the limit equals $0$. However I am not sure how to prove the limit exists? Do we have to do a $\epsilon -\delta$ proof?

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  • $\begingroup$ Yeah it is equal to zero and I think that's a good proof. $\endgroup$ – Shahar May 11 '14 at 22:35
  • $\begingroup$ If you can already do a proof using a theorem and still wants an even elementary proof, I would say refer how the theorem is proved and repeat the same step for this specific application of theorem. $\endgroup$ – jdoicj May 13 '14 at 17:08
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You should reread the version of L'Hôpital you have learned.

It is a theorem that guarantees the existence of the limit if it is applicable. Therefore, you do not have to give any other proof. However, you do have to check that the conditions of L'Hôpitals theorem are fulfilled before you apply it.

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$$ 0\ <\ \left\vert\,x\ln\left(x\right)\,\right\vert\ \leq\ \left\vert\,x\left(x - 1\right)\right\vert $$

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