Show, using L'Hopitals rule, that \[\lim_{x \rightarrow 0} x^{a}lnx = 0\] for a>0 I started by doing the first derivitive but it didn't help much and doing more derivitives just made it worse, how should i continue? \[\lim_{x \rightarrow 0} \frac{ d }{dx} x ^{a}lnx = \lim_{x \rightarrow 0} x^{a-1}(alnx+1)\]

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Show, using L'Hopitals rule, that \[\lim_{x \rightarrow 0} x^{a}lnx = 0\] for a>0 I started by doing the first derivitive but it didn't help much and doing more derivitives just made it worse, how should i continue? \[\lim_{x \rightarrow 0} \frac{ d }{dx} x ^{a}lnx = \lim_{x \rightarrow 0} x^{a-1}(alnx+1)\]

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remember that you can only apply l'hospital rule on specific forms (the only one i remember i 0/0 at the moment.) remember that \[x^a = \ln (e ^{x^a})\]
I forgot that, so now I should just derivate? So: \[\lim_{x \rightarrow 0} \frac{ d }{ dx} \ln(e ^{x ^{a}})lnx=\lim_{x \rightarrow 0}\frac{ ax ^{a} lnx \ln(e ^{x ^{a}}) }{ x } \] And that is 0/0, and to get the answer should I just continue to derivate?
It does get really messy to derivate more.. or maybe I shouldn't derivate further?

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Ohh i see, thanks :)

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