Exploring the Limit Problem: A Unique Approach Using Maclaurin Series

Calculus enthusiasts and students often encounter problems requiring the evaluation of limits that involve trigonometric functions. One such intriguing example is the limit problem presented above. In this article, we will delve into the detailed solution of the limit problem without relying on L'Hostpital's rule, but rather using the Maclaurin series expansion technique.

Introduction to the Limit Problem

The problem in question is the evaluation of the following limit:

$$ L lim_{x to 0} frac{arcsin x - x}{{sin}^3 x} $$

Evaluating the Limit Using Maclaurin Series

Maclaurin series are a powerful tool in calculus, providing us with expansions of functions around zero. These expansions are particularly useful for evaluating limits where direct substitution does not yield a determinate form.

Step 1: Expressing Arcsin(x) and Sin(x)

We will first express the Maclaurin series for (arcsin x - x) and (sin x^3). The Maclaurin series for (arcsin x) is given by:

$$ arcsin x x frac{x^3}{6} O(x^5) $$

where (O(x^5)) represents higher-order terms that become insignificant as (x) approaches 0. For (sin x^3), we have:

$$ sin x^3 x^3 - frac{(x^3)^3}{6} O(x^9) x^3 - frac{x^9}{6} O(x^9) $$

Step 2: Substitution and Simplification

Substituting these series into the limit expression, we get:

$$ L lim_{x to 0} frac{arcsin x - x}{x^3} cdot left( frac{x}{sin x} right)^3 cdot left( frac{sin x}{sin(sin x)} right)^3 $$

Breaking this into separate terms, we have:

$$ L lim_{x to 0} frac{arcsin x - x}{x^3} cdot lim_{x to 0} left( frac{x}{sin x} right)^3 cdot lim_{x to 0} left( frac{sin x}{sin(sin x)} right)^3 $$

Step 3: Evaluating Each Limit

We now evaluate each of these limits one by one. Starting with the first term:

$$ L_1 lim_{x to 0} frac{arcsin x - x}{x^3} lim_{x to 0} frac{x - left( x - frac{x^3}{6} right)}{x^3} lim_{x to 0} frac{x - x frac{x^3}{6}}{x^3} frac{1}{6} $$

The second term simplifies as follows:

$$ L_2 lim_{x to 0} left( frac{x}{sin x} right)^3 1 $$

The third term can be evaluated using the equivalence of functions near zero:

$$ L_3 lim_{x to 0} left( frac{sin x}{sin(sin x)} right)^3 1 $$

Final Result

Combining these results, we obtain the final answer:

$$ L L_1 cdot L_2 cdot L_3 frac{1}{6} cdot 1 cdot 1 frac{1}{6} $$

Conclusion

Through the use of Maclaurin series and careful limit evaluation, we have demonstrated that the given limit evaluates to (frac{1}{6}). This method offers a unique approach to solving the limit problem, avoiding the need for L'Hopital's rule. As always, understanding the underlying principles and techniques is key to mastering advanced mathematical concepts.