Answer to Question #205017 in Calculus for Zakkir

Question

Find the value of a and b, if

Lim_x→∞ [x(1 + acosx) - b sin x]/x³ = 1

Answer 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 .