lim-x-0-x-1-a-cos-x-bsin-x-x-5-1-find-a-amp-b-

Question Number 105410 by john santu last updated on 28/Jul/20

\lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{x^{5}} = 1

f i n d a & b

Answered by bobhans last updated on 29/Jul/20

\lim_{x \to 0} \frac{x (1 + a (1 - \frac{1}{2} x^{2} + \frac{1}{24} x^{4})) - b (x - \frac{1}{6} x^{3} + \frac{x^{5}}{120})}{x^{5}} = 1

\lim_{x \to 0} \frac{x (1 + a - \frac{1}{2} {a x}^{2} + \frac{1}{24} {a x}^{4}) - (b x - \frac{1}{6} {b x}^{3} + \frac{b}{120} x^{5})}{x^{5}} = 1

\lim_{x \to 0} \frac{(1 + a - b) x + (\frac{1}{6} b - \frac{1}{2} a) x^{3} + (\frac{1}{24} a - \frac{1}{120} b) x^{5}}{x^{5}} = 1

(1) 1 + a - b = 0; a - b = - 1

(2) \frac{1}{6} b - \frac{1}{2} a = 0; b = 3 a

(3) \frac{1}{24} a - \frac{1}{120} b = 1; a = \frac{1}{5} b

t h e q u e s t i o n i n c o n s i s t e n t

Commented by 1549442205PVT last updated on 29/Jul/20

I think that in the question need to give

three unknown

Answered by Dwaipayan Shikari last updated on 29/Jul/20

\lim_{x \to 0} \frac{x (1 + a c o s x) - b s i n x}{x^{5}} = \frac{x a (- s i n x) + a c o s x - b c o s x}{5 x^{4}}

= \frac{x a (- c o s x) + a (- s i n x) + a (- s i n x) + b s i n x}{20 x^{3}}

= \frac{x a s i n x + a (- c o s x) + a (- c o s x) + a (- c o s x) + b c o s x}{60 x^{2}}

= \frac{x a c o s x + a s i n x + 3 a s i n x - b s i n x}{120 x} o r \frac{x a (- s i n x) + a c o s x + 4 a c o s x - b c o s x}{120}

= \frac{a c o s x}{120} + \frac{a x}{120 x} + \frac{3 a x}{120 x} - \frac{b x}{120 x} o r \frac{5 a c o s x - b c o s x}{120}

= \frac{a}{24} - \frac{b}{120}

\Rightarrow \frac{a}{24} - \frac{b}{120} = 1

5 a - b = 120 \to (1)

\lim_{x \to 0} \frac{x (1 + a c o s x) - b s i n x}{x^{5}} = 1

1 + a c o s x - b = 0

1 + a = b \to (2)

5 a - 1 - a = 120

a = \frac{121}{4}

b = \frac{125}{4}

Commented by 1549442205PVT last updated on 30/Jul/20

“ \lim \frac{x (1 + a c o s x) - b s i n x}{x^{5}} = 1

1 + a c o s x - b = 0^{″} (1)

I think that ocurred a mistake at this

place . It is only possible

1 + a c o s x - b = 0 \Rightarrow \lim \frac{x (1 + a c o s x) - b s i n x}{x^{5}} = 1

and it is impossible

\Rightarrow \lim \frac{x (1 + a c o s x) - b s i n x}{x^{5}} = 1 \Rightarrow 1 + a c o s x - b = 0

it is also like as 1 + acosx - kb = 0

\Rightarrow \lim \frac{x (1 + a c o s x) - b s i n x}{x^{5}} = 1

If (1) is true then 1 + acosx - kb = 0

is also true . Sir should see once again

{sir}^{'} argument