Differential Calculus
Plz 12 question
2 Answers
srishti
((1-cos2x)sin5x)/((x^2)(sin3x) where lim tends to 0
1-cos2x=2(sin^x)
sin^x/(x^2)=1
hence,
2(((sin^x)^2)/x^2)*(sin5x/sin3x)
2*(sin5x)/(sin3x)
using l’ hopital rule
2*(cos5x)*5)/(cos3x).3)
putting the limits
2*5/3
10/3
srishti
e^(x^2)-cosx/x^2
using l hopital rule
2xe^2+sinx/2x
4x^2e^x+cosx/2
1/2
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