Q 10 3 is (1) 1/8 (2) 0 (3) 1/32 (4) - Maths - Continuity and Differentiability

Question

Q 10 
3 is (1) 1/8 (2) 0 (3) 1/32 (4) infinity

Namita_2 · Accepted Answer

Dear Student
If your question is

limx→π21-tanx21-sinx1+tanx2π-2x3Then the answer isFirst we have to simplify 1-tanx21+tanx2Multiplying both numerator and denominator with cosx2 we getcosx2-sinx2cosx2+sinx2now rationalizing the by multiplying with cosx2-sinx2we get1-tanx21+tanx2=cosx2-sinx2cosx2+sinx2×cosx2-sinx2cosx2-sinx2=cosx2-sinx22cos2x2-sin2x2=cos2x2+sin2x2-2sinx2cosx2cosx=1-sinxcosxThus the question simplifies to1-sinx2cosxπ-2x3Let π-2x=tx→π2,t→0also x=π2-t2⇒cosx=cosπ2-t2=sint2=2sint4cost4 Also 1-sinx=1-sinπ2-t2=1-cost2=2sin2 t4⇒1-sinx 2=4sin4t4Tha above equation transforms tolimt→04sin4t42sint4cost4t3⇒2limt→0sin3t4cost4t3⇒2limt→0sin3t4t3                as t→0 cost4=cos0=1    ⇒2143⇒264=132  
Regards

Q.10 3 is (1) 1/8 (2) 0 (3) 1/32 (4) infinity