Question

find the limit of this.​

Answer 1

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\lim _{x\to 0}\dfrac{x(1 + acosx) - bsinx}{ {x}^{3} } = 1$

We know,

$\rm :\longmapsto\:cosx = 1 - \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{4} }{4!} - \dfrac{ {x}^{6} }{6!} - - -$

$\rm :\longmapsto\:acosx = a - \dfrac{ {ax}^{2} }{2!} + \dfrac{ {ax}^{4} }{4!} - \dfrac{ {ax}^{6} }{6!} - - -$

$\rm :\longmapsto\:1 + acosx =1 + a - \dfrac{ {ax}^{2} }{2!} + \dfrac{ {ax}^{4} }{4!} - \dfrac{ {ax}^{6} }{6!} - - -$

and

$\rm :\longmapsto\:sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} - \dfrac{ {x}^{7} }{7!} - - -$

$\rm :\longmapsto\:bsinx = bx - \dfrac{ {bx}^{3} }{3!} + \dfrac{ {bx}^{5} }{5!} - \dfrac{ {bx}^{7} }{7!} - - -$

On substituting all these values in given, we get

$\rm \:\displaystyle\lim _{x\to 0}\dfrac{x\bigg(1+a -\dfrac{{ax}^{2} }{2!}+\dfrac{{ax}^{4} }{4!} + - - \bigg) -\bigg(bx -\dfrac{{bx}^{3}}{3!}+ \dfrac{ {bx}^{5} }{5!} + - - - \bigg) }{ {x}^{3} } = 1$

$\rm \:\displaystyle\lim _{x\to 0}\dfrac{\bigg(1+a -\dfrac{{ax}^{2} }{2!}+\dfrac{{ax}^{4} }{4!} + - - \bigg) -\bigg(b -\dfrac{{bx}^{2}}{3!}+ \dfrac{ {bx}^{4} }{5!} + - - - \bigg) }{ {x}^{2} } = 1$

$\: \rm\displaystyle\lim _{x\to 0}\dfrac{(1 + a - b) + {x}^{2} \bigg(\dfrac{b}{3!} - \dfrac{a}{2!} \bigg) + {x}^{4}\bigg(\dfrac{a}{4!} - \dfrac{b}{5!} \bigg) + - - - }{ {x}^{2} } = 1$

For limit to be finite,

$\rm :\longmapsto\:1 + a - b = 0$

$\bf\implies \:b = 1 + a - - - (1)$

So, given limit reduced to

$\: \rm\displaystyle\lim _{x\to 0}\dfrac{{x}^{2} \bigg(\dfrac{b}{3!} - \dfrac{a}{2!} \bigg) + {x}^{4}\bigg(\dfrac{a}{4!} - \dfrac{b}{5!} \bigg) + - - - }{ {x}^{2} } = 1$

$\: \rm\displaystyle\lim _{x\to 0}\bigg(\dfrac{b}{3!} - \dfrac{a}{2!} \bigg) + \lim _{x\to 0}{x}^{2}\bigg(\dfrac{a}{4!} - \dfrac{b}{5!} \bigg) + - - - = 1$

$\rm :\longmapsto\:\dfrac{b}{3!} - \dfrac{a}{2!} = 1$

$\rm :\longmapsto\:\dfrac{b}{3 \times 2 \times 1} - \dfrac{a}{2 \times 1} = 1$

$\rm :\longmapsto\:\dfrac{b}{6} - \dfrac{a}{2} = 1$

$\rm :\longmapsto\:\dfrac{b - 3a}{6} = 1$

$\rm :\longmapsto\:b - 3a = 6$

$\rm :\longmapsto\:1 + a - 3a = 6 \: \: \: \: \: \: \: \: \: \{ \: using \: (1) \: \}$

$\rm :\longmapsto\:1 - 2a = 6$

$\rm :\longmapsto\:- 2a = 6 - 1$

$\rm :\longmapsto\:- 2a = 5$

$\bf\implies \: a = \: - \: \dfrac{5}{2}$

On substituting the value of a in equation (1) we get

$\rm :\longmapsto\:b = 1 - \dfrac{5}{2}$

$\rm :\longmapsto\:b = \dfrac{2 - 5}{2}$

$\bf\implies \:b \: = \: - \: \dfrac{3}{2}$