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frx
Group Title
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?
\[\lim_{x \rightarrow 0} \frac{ d }{dx} x ^{a}lnx = \lim_{x \rightarrow 0} x^{a1}(alnx+1)\]
 2 years ago
 2 years ago
frx Group Title
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? \[\lim_{x \rightarrow 0} \frac{ d }{dx} x ^{a}lnx = \lim_{x \rightarrow 0} x^{a1}(alnx+1)\]
 2 years ago
 2 years ago

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alexandercpark Group TitleBest ResponseYou've already chosen the best response.0
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 \[x^a = \ln (e ^{x^a})\]
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
It does get really messy to derivate more.. or maybe I shouldn't derivate further?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1351678852308:dw
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Ohh i see, thanks :)
 2 years ago
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