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Mathematics 19 Online
OpenStudy (anonymous):

Evaluate limx->0 (2x^2+x)/sinx do not use L'Hospital's rule

OpenStudy (anonymous):

graph this

OpenStudy (anonymous):

take x common from numerator x(2x+1)/sinx =(x/sinx)*(2x+1) now apply limit ...and usind sandwich theorem sinx/x=1 so x/sinx=1 answer =1

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