Answered: "limx→0x(1+acos⁡x)−bsin⁡xx3=1, how to find the constants a,b?"

hadejada7x

Answered question

2022-01-16

lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{x^{3}} = 1

, how to find the constants a,b?

Vivian Soares

Beginner2022-01-17Added 36 answers

If you can use Taylor series, the numerator is

x (1 + a - \frac{a x^{2}}{2}) - b (x - \frac{x^{3}}{6}) + o (x^{3}) = (1 + a - b) x + (b - 3 a) \frac{x^{3}}{6} + o (x^{3})

Solving

1 + a - b = 0, b - 3 a = 6

gives

a = - \frac{52}{,} b = - \frac{32}{}

kaluitagf

Beginner2022-01-18Added 38 answers

Apply LHopitals rule again:

lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{x^{3}} \overset{L H R}{=} lim_{x \to 0} \frac{1 + (a - b) \cos x - a x \sin x}{3 x^{2}}

\overset{L H R}{=} lim_{x \to 0} \frac{(b - 2 a) \sin x - a x \cos x}{6 x}

and recall that

lim_{x \to 0} \frac{\sin x}{x} = 1

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?