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By Wolfram $$\lim_{x\to 0^+}\frac{\ln(x)}{x}=-\infty $$ whereas by l'hopital it makes $$\lim_{x\to 0^+}\frac{1}{x}=+\infty,$$ why l'Hopital doesn't work here ?

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  • $\begingroup$ This limit does not exist. $\endgroup$ – Jed Nov 8 '15 at 22:33
  • $\begingroup$ I would advise you not to use l'Hospital before trying to get the limit with another method. Firstly because l'Hospital only applies in specific case (a necessary but not sufficient condition is that the limit is of the type $0/0$ or $\pm\infty/\pm\infty$) and because it sometimes (often, actually) implies a lot of calculus, so it is worth choosing a method allowing you to avoid l'Hospital, even if it seems (and it is) a very powerful rule. $\endgroup$ – MoebiusCorzer Nov 8 '15 at 23:07
  • $\begingroup$ @MoebiusCorzer I don't even think L'Hopital is that powerful. In some cases yes, in most cases no. Usually it is a method that any mildly engaged student can toss at a problem without much thought involved, leading to an answer that agrees with the back of the book. Meanwhile nothing has been learned. $\endgroup$ – zhw. Nov 8 '15 at 23:25
  • $\begingroup$ @MoebiusCorzer One other thing. LHR is applicable in the case the denominator is $\pm \infty.$ We don't need to know what the numerator is doing. $\endgroup$ – zhw. Nov 8 '15 at 23:27
  • $\begingroup$ @zhw I agree with you, but most students don't understand much the "methods" to find limits, L'Hospital included. I said it is powerful because it is useful to compute particular limits easily. The way to learn by using methods is to understand why the method work, whatever it is. $\endgroup$ – MoebiusCorzer Nov 8 '15 at 23:28
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It doesn't work since it's of the form $+\infty \times -\infty $ and it's not an indetermination. Therefore, you can't use l'Hopital rule here. You will have more information on the utilisation of l'Hopital Rule here.

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