Introduction

In this article, we are going to solve a complex limit problem using the Taylor series expansion. The objective is to find the values of a and b for which the limit

(lim_{xto0}frac{x(1-acos x) - bsin x}{x^3} 1)

1. Understanding the Limit Function

To solve this problem, we need to simplify the expression inside the limit and apply the Taylor series expansion for (cos x) and (sin x) around (x 0).

1.1 Taylor Series Expansion

The Taylor series expansions for (cos x) and (sin x) around (x 0) are given by:

(cos x approx 1 - frac{x^2}{2} O(x^4)) (sin x approx x - frac{x^3}{6} O(x^5))

1.2 Substituting the Series into the Limit

Substituting these approximations into the original limit expression, we get:

[x(1-acos x) - bsin x approx xleft(1-aleft(1-frac{x^2}{2}right) O(x^4)right) - bleft(x-frac{x^3}{6} O(x^5)right)]

Expanding and simplifying:

[x(1-acos x) - bsin x approx xleft(1 - a frac{ax^2}{2} O(x^4)right) - bx frac{bx^3}{6} O(x^5)]

This further simplifies to:

[x(1 - a) frac{ax^3}{2} - bx frac{bx^3}{6} O(x^5) x(1 - a - b) left(frac{a}{2} frac{b}{6}right)x^3 O(x^5)]

Therefore, the limit becomes:

[lim_{xto0}frac{x(1 - a - b) left(frac{a}{2} frac{b}{6}right)x^3 O(x^5)}{x^3} 1]

1.3 Solving for (a) and (b)

For the limit to be equal to 1, the coefficient of (x^3) in the numerator must be 1, and the rest of the terms must approach zero as (xto 0).

(1 - a - b 0) implies (b 1 - a) (frac{a}{2} frac{b}{6} 1)

Substituting (b 1 - a) into the second equation:

[frac{a}{2} frac{1 - a}{6} 1]

Combining the terms and solving for (a) :

[frac{3a 1 - a}{6} 1] [frac{2a 1}{6} 1] [2a 1 6] [2a 5] [a -frac{5}{3}]

Now substituting (a) back to find (b) :

(b 1 - a 1 - (-frac{5}{3}) frac{2}{3})

Thus, the values of (a) and (b) are:

(a -frac{5}{3}, b -frac{2}{3})

2. An Alternate Approach

For clarity, let's consider an alternative approach by expressing (cos x) as a Taylor series and simplifying the numerator.

2.1 Simplifying Using Taylor Series

Assuming:

(cos x 1 - frac{x^2}{2!} frac{x^4}{4!} - ...)

Let the numerator be:

(x(1 - acos x) x(1 - a frac{ax^2}{2} - frac{ax^4}{4!} ... - bx frac{bx^3}{6} - ...)

Since the problem demands the numerator to be approximately equal to (x^3), we set (b 0) and observe the constant term and linear terms.

The constant term should be 0 and the coefficient of the linear term should also be 0.

2.2 Solving for (a)

Setting the constant and linear terms to 0:

Constant term: (1 - a 0) implies (a 1) Linear term: (-a 0) implies (a 0)

This leads to a contradiction, which means there is no unique solution under these conditions.

In conclusion, we have two distinct methods providing a clear understanding of how to solve the problem and obtain the values of (a) and (b).