# Answer in Calculus for Zakkir #205017

Find the value of *a *and *b*, if

Lim_{x}→∞ [x(1 + *a*cos*x*) – *b* sin *x*]/x^{3 } = 1

“lim_{x to infty}frac{x(1 + acos(x)) – b sin (x)}{x^3}” **(**“frac{infty}{infty}”**) form**

since above form is an indeterminant form. therefore using** L’Hopital rule **

“implies” “lim_{x to infty}frac{frac{d}{dx}[x(1 + acos(x) – b sin(x)]}{frac{d}{dx}x^3}”

“implies” “lim_{x to infty}frac{frac{d}{dx}(x + axcos(x)) – frac{d}{dx}b sin(x)]}{frac{d}{dx}x^3}”

“implies” “lim_{x to infty}frac{(1 + acos(x)-axsin(x)) -b cos(x)}{3x^2}” **(**“frac{infty}{infty}”**) form**

again applying **L’Hopital rule ,**

“implies” “lim_{x to infty}frac{frac{d}{dx}[(1 + acos(x)-axsin(x)) -b cos(x)]}{frac{d}{dx}3x^2}”

“implies” “lim_{x to infty}frac{[(0 – asin(x)-axcos(x)-asin(x)) + b sin(x)]}{6x}”

“implies” “lim_{x to infty}frac{(asin(x)-axcos(x)-asin(x) + b sin(x)}{6x}” **(**“frac{infty}{infty}”**) form**

again applying **L’Hopital rule,**

“implies” “lim_{x to infty}frac{frac{d}{dx}(asin(x)-axcos(x)-asin(x) + b sin(x)}{frac{d}{dx}6x}”

“implies” “lim_{x to infty}frac{(acos(x)+axsin(x)-acos(x)-acos(x) + b cos(x)}{6}” “to” (1)

we get , “lim_{x to infty}frac{x(1 + acos(x)) – b sin (x)}{x^3}” “neq” 1 for any a and b because each term of numerator contain either cos(x) or sin(x) which always oscillate between 1 and -1.

if we take a = 0 in (1) then it becoms “frac{b cos(x) }{6}” which can never tends to 1 as x “to” “infty”

Hence, the given question is not Right .

Dear Raghav Sood, thank you for leaving a feedback. A solution of the question uses another approach.