Determine the limx→1lnxsinπx. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.
limx→1lnxsinπx=ln(1)sin[(π)(1)]=0sinπ=00 Indeterminate form
Thus, by applying L'Hospital's Rule...
limx→1lnxsinπx=limx→1xddx(lnx)ddxsinπ(x)=limx→11xcosπx⋅π=limx→11πxcos(πx)=1π(1)cos(π(1))=1πcos(π)=1π(−1)=−1π
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