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Lt x-0 x(1+acosx) - bsinx/x^3=1Find the values of a and b

Lt x-0 x(1+acosx) - bsinx/x^3=1Find the values of a and b

Grade:11

1 Answers

Arun
25750 Points
6 years ago
We have lim x->0 [x(1-acosx)+bsinx]/x^3=1/3 
but lim x->0 [x(1-acosx)+bsinx]/x^3 has the 
form 0/0 so from L' Hospital's Rule 
taking derivatives we havelim x->0 [1-acosx+axsinx+bcosx]/3x^2 
=1/3 (1) 
Putting g(x)= [1-acosx+axsinx+bcosx]/3x^2 
(3x^2)g(x)= [1-acosx+axsinx+bcosx] 
Taking limx-->0 of both sides we have 
0*1/3=1-a+b-->b=a-1(2) 
(1) because of (2) becomes limx->0 
[1-acosx+axsinx+acosx-cosx] 
/3x^2=1/3 
lmx->0[1+axsinx-cosx]/3x^2=1/3 
L'Hospital's rule again 
lim x-->0[asinx+axcosx+sinx]/6x 
=1/3 
L'Hospital's rule again 
lim x-->0 [acosx+acosx-axsinx+cosx]/6 
=1/3 
lim x-->0 [2acosx-axsinx+cosx]/6=1/3 
Should be 
lim x-->0 [2acosx-axsinx+cosx]=2 
2a +1 =2-->a=1/2 and b=-1/2

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