Question 28 of 401MediumMCQ
x→0limcscx(2cos2x+3cosx−cos2x+sinx+4) is: Solutions · 1
Official solution
x→0limcscx(2cos2x+3cosx−cos2x+sinx+4)x→0lim(2cos2x+3cosx+cos2x+sinx+4)csc(cos2x+3cosx−sinx−4)x→0limsinx1(2cos2x+3cosx+cos2x+sinx+4)(cos2x+3cosx−4)−sinxx→0limsinx(2cos2x+3cosx+cos2x+sinx+4)(cosx+4)(cosx−1)−sinxx→0lim2sin2xcos2x(2cos2x+3cosx+cos2x+sinx+4)−2sin22x(cosx+4)−2sin2xcos2xx→0limcos2x(2cos2x+3cosx+cos2x+sinx+4)−(sinx2x(cosx+4)+cos2x)−251 Answer the question to unlock