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Question

# limx→π/2tan2x[(2sin2x+3sinx+4)1/2−(sin2x+6sinx+2)1/2] is 1m Find m

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Solution

## Put x=π2 in tan2x[(2sin2x+3sinx+4)12−(sin2x+6sinx+2)12]=tan2π2⎡⎣(2sin2π2+3sinπ2+4)12−(sin2π2+6sinπ2+2)12⎤⎦=∞[(2+3+4)12−(1+6+2)12]=∞[(9)12−(9)12]=∞×0 is an indeterminate formConsider tan2x[(2sin2x+3sinx+4)12−(sin2x+6sinx+2)12]=tan2x⎡⎢ ⎢ ⎢ ⎢⎣((2sin2x+3sinx+4)12−(sin2x+6sinx+2)12)((2sin2x+3sinx+4)12+(sin2x+6sinx+2)12)(2sin2x+3sinx+4)12+(sin2x+6sinx+2)12⎤⎥ ⎥ ⎥ ⎥⎦=tan2x2sin2x+3sinx+4−sin2x−6sinx−2√2sin2x+3sinx+4+√sin2x+6sinx+2=tan2xsin2x−3sinx+2√2sin2x+3sinx+4+√sin2x+6sinx+2limx→π2tan2x[(2sin2x+3sinx+4)12−(sin2x+6sinx+2)12]=limx→π2tan2xsin2x−3sinx+2√2sin2x+3sinx+4+√sin2x+6sinx+2=limx→π2tan2xsin2x−3sinx+2√2sin2π2+3sinπ2+4+√sin2π2+6sinπ2+2=limx→π2tan2xsin2x−3sinx+2√2+3+4+√1+6+2=limx→π2tan2xsin2x−2sinx−sinx+2√9+√9=limx→π2tan2xsinx(sinx−2)−1(sinx−2)√9+√9=limx→π2tan2x(sinx−2)(sinx−1)3+3=limx→π2tan2x(sinx−2)(sinx−1)6Put x=π2+h for h→0=limh→0tan2(π2+h)(sin(π2+h)−2)1(sin(π2+h)−1)6=limh→0cot2h(cosh−2)(cosh−1)6=limh→0(cosh−2)(cosh−1)6tan2h=limh→0(cosh−2)(cosh−1)6h2tan2hh2=limh→0(cosh−2)(cosh−1)6h2By using L'Hospitals Rule we get=limh→0(cosh−2)(−sinh)+(cosh−1)(−sinh)12h=limh→0−sinh(cosh−2+cosh−1)12h=limh→0−sinh(2cosh−3)12h=limh→0−(2cosh−3)12=−212+312=112⇒1m=112∴m=12

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