Recent content by fxo

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    L'Hospital's Rule for Limit of x^a ln(x) as x Approaches 0 (a>0)

    By setting ##x=e^t## in the first one I got$$ \lim_{t\rightarrow\, 0} te^t^a $$ I'm starting to see that it's close to the form in b but not exactly, should I use L'Hospitals rule from here to show that as this goes to infinity the limit goes to zero?
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    L'Hospital's Rule for Limit of x^a ln(x) as x Approaches 0 (a>0)

    ##\lim_{e^t\rightarrow 0}\frac {\ln (e^t}{\frac 1 {e^t^a}}## I don't really get it what's the connection to the new limit, do they want me to change the e^x to x=ln(t) and get |x|^a * t instead?
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    L'Hospital's Rule for Limit of x^a ln(x) as x Approaches 0 (a>0)

    $$ \lim_{ln(t)\rightarrow 0}\frac {\ln (ln(t))}{\frac 1 {ln(t)^a}}$$ So I get $$ ln(ln(t))ln^a(t)$$ But that expression don't tell me much all I know is that it's root should be e.
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    L'Hospital's Rule for Limit of x^a ln(x) as x Approaches 0 (a>0)

    Oh sorry you're right I totally missed that a>0. I get it then. Can I ask a follow-up question based on this one? By setting x=ln(t) in the equation above(the one I asked for help on first) show that: $$ \lim_{x\rightarrow -∞} |x|^a e^x = 0$$ This one really bugs my mind I just don't get it.
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    L'Hospital's Rule for Limit of x^a ln(x) as x Approaches 0 (a>0)

    Homework Statement Use L'Hospitals rule to show that lim x->0 x^a ln(x) = 0 I don't know how to solve this. I guess the first thing to do is to transform it in some way so that one can use L'Hospitals rule, but I don't know how. Thank you! EDIT: a>0 It's not a differential equation as...