Answer to Question #206263 in Calculus for Nikhil rawat

Question

Find the value of a and b if

lim x(1+acosx)-bsinx/ (x^3)=1

x→∞

Answer 1

Solution:-

there is error in question as i got when I solved the full question that is it comes out to be infinite , the limit should tend to zero instead of infinite to get the result , so i am solving the problem by taking limit as tends to zero.

$lim_{x \rightarrow 0} {x(1+acosx)-bsinx\over (x^3)}=1$

as we see it is ${0 \over 0}$ type of limit

using L' Hospital rule

we get

=$lim_{x \rightarrow 0} {1+acosx-axsinx-bcosx\over (3x^2)}=1$

=$lim_{x \rightarrow 0} {1+(a-b)cosx-axsinx\over (3x^2)}=1$

we see that limit exist when a-b=-1 , the $0\over 0$ exist with this condition only

To use L'Hospital , we see a-b has to be -1

$\therefore$ $a-b =-1$ ............................................................(1)

Applying L'Hospital again ,

= $lim_{x \rightarrow 0} {-(a-b)sinx-asinx-axcosx\over (6x)}=1$

this is is in form of ${0 \over 0}$ , using L'Hospital yet again ,

= $lim_{x \rightarrow 0} {(b-2a)cosx+axsinx-acosx\over (6)}=1$

=$lim_{x \rightarrow 0} {(b-3a)cosx+axsinx\over (6)}=1$ { sin(0)=0 , cos(0)=1}

$\implies$

when we put x=0 , we get cosx =1, sinx=0, and putting the limit equals to 1 we get the following result

b-3a=6 ...........................................................................(2)

solving (1) and (2) we get

a=$-5\over 2$

b=$-3\over 2$