Lt x-0 x(1+acosx) - bsinx/x^3=1Find the values of a and b
YASH AHUJA , 7 Years ago
Grade 11
Differential Calculus> Lt x-0 x(1+acosx) - bsinx/x^3=1Find the v...
1 Answers
Arun
Last Activity: 7 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
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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