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Mathematics 12 Online
OpenStudy (anonymous):

Find the limit as x tends to 0+: lim(sqrt(x)*ln(x))

OpenStudy (dumbcow):

not sure if this is best answer, but if you assume 0^0 = 1 , then the limit is 0 \[\lim_{x \rightarrow 0+}\sqrt{x} \ln x = \ln e^{\lim_{x \rightarrow 0+}\sqrt{x} \ln x} = \ln \lim_{x \rightarrow 0+}x^{\sqrt{x}} = \ln 1 = 0\]

OpenStudy (anonymous):

Hmm. Not sure how you got from the 2nd to the 3rd stage.

OpenStudy (anonymous):

Although that is the right answer

OpenStudy (anonymous):

I think I get it - does the \[e ^{\ln x}\] cancel out?

OpenStudy (dumbcow):

oh sorry, yes e^lnx = x

OpenStudy (dumbcow):

\[e^{\ln x * \sqrt{x}} = (e^{\ln x})^{\sqrt{x}} = x^{\sqrt{x}}\]

OpenStudy (anonymous):

Thanks! Here's your hard-earned "best response"

OpenStudy (dumbcow):

haha thx

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