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Find the limit as x tends to 0+: lim(sqrt(x)*ln(x))
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not sure if this is best answer, but if you assume 0^0 = 1 , then the limit is 0 \[\lim_{x \rightarrow 0+}\sqrt{x} \ln x = \ln e^{\lim_{x \rightarrow 0+}\sqrt{x} \ln x} = \ln \lim_{x \rightarrow 0+}x^{\sqrt{x}} = \ln 1 = 0\]
Hmm. Not sure how you got from the 2nd to the 3rd stage.
Although that is the right answer
I think I get it - does the \[e ^{\ln x}\] cancel out?
oh sorry, yes e^lnx = x
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\[e^{\ln x * \sqrt{x}} = (e^{\ln x})^{\sqrt{x}} = x^{\sqrt{x}}\]
Thanks! Here's your hard-earned "best response"
haha thx
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