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Calculus1
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Limit as x approach 0+ of √x lnx
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can you use l'hopital?
is it \[\sqrt{x}\times \ln (x)\] or \[\sqrt{x \times \ln x}\]
usual gimmick it to rewrite as \[\frac{\ln(x)}{\frac{1}{\sqrt{x}}}\] and then use l'hopital
\[\sqrt{x} lnx \]
I've tried with L'hospital but am getting stuck in the work
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When I use L'hopital I get \[\lim_{x \rightarrow 0+} = -2\times \sqrt{x} = 0\] and when I graph it on my TI I get the same
thank you :)
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