I'm having trouble with finding the limit of (sinx)(lnsinx) as x approaches 0 from the positive side. Any suggestions? Thanks!
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Why not try L'Hospital's rule? $\displaystyle sin(x) log(sin(x)) = \frac{log(sin(x))}{\frac{1}{sinx}} $. That limit has the indeterminate form $\displaystyle \frac{\infty}{\infty} $.
What he said
For x 'small enough' is $\displaystyle \sin x \approx x$ so that is... $\displaystyle \lim_{ x \rightarrow 0+} \sin x \cdot \ln \sin x = \lim_{ x \rightarrow 0+} x \cdot \ln x = 0$ Kind regards $\displaystyle \chi$ $\displaystyle \sigma$
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