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OpenStudy (anonymous):
lim[(cosx+1)/(xsins)] (for lim x to π)
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OpenStudy (anonymous):
use l hopital to solve as it is indeterminate form it will be lim of (-sinx)/(sinx-xcos(x)), if you replace pi inside it will give 0/pi=0
OpenStudy (anonymous):
sorry denominator is sin(x)+xcos(x) so 0/-pi = 0 still
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