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Evaluate the following limit lim as x approaches 0 from the right (1/x) integral [1,(1-x)] of t^(ln(1-t)) dt
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L'hopitals rule!!
lim = 1 case for x=.0001 http://www.wolframalpha.com/input/?i=integrate+t%5Elog%281-t%29+dt+from+0.9999+to+1
\[\lim_{x\to0^+}\frac{\int_1^{1-x}t^{\ln(1-t)}~dt}{x}\] Upon substituting \(x=0\), you have \[\lim_{x\to0^+}\frac{\int_1^{1}t^{\ln(1-t)}~dt}{0}=\frac{0}{0}\] Apply L'Hopital's rule. \[\lim_{x\to0^+}\frac{\frac{d}{dx}\int_1^{1-x}t^{\ln(1-t)}~dt}{\frac{d}{dx}x}\] By the FTC, you have \[\lim_{x\to0^+}\frac{-(1-x)^{\ln(1-(1-x))}}{1}\]
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