Evaluating the Limit of (frac{e^x - e^{sin x}}{x - sin x}) at (x 0) Using L'H?pital's Rule and Taylor Series

Introduction:

Understanding different methods to evaluate limits, particularly those involving indeterminate forms, is fundamental in calculus. This article will explore the evaluation of the following limit using both L'H?pital's Rule and Taylor Series expansions:

(lim_{xto 0} frac{e^x - e^{sin x}}{x - sin x})

L'H?pital's Rule

Since both the numerator and the denominator approach 0 as (x) approaches 0, L'H?pital's Rule can be applied. The rule states that if (lim_{xto c} frac{f(x)}{g(x)}) is of the form (frac{0}{0}) or (frac{infty}{infty}), then:

(lim_{xto c} frac{f(x)}{g(x)} lim_{xto c} frac{f'(x)}{g'(x)})

Derivatives of the Numerator

The numerator of the limit is:

(e^x - e^{sin x})

The derivative of (e^x) is (e^x) and the derivative of (e^{sin x}) using the chain rule is (e^{sin x} cdot cos x). Therefore, the derivative of the numerator is:

(e^x - e^{sin x} cos x)

Derivatives of the Denominator

The denominator of the limit is:

(x - sin x)

The derivative of (x) is 1, and the derivative of (sin x) is (cos x). Therefore, the derivative of the denominator is:

(1 - cos x)

Status after applying L'H?pital's Rule:

(lim_{xto 0} frac{e^x - e^{sin x} cos x}{1 - cos x})

When we substitute (x 0), the numerator becomes:

(e^0 - e^{sin 0} cos 0 1 - 1 0)

And the denominator becomes:

(1 - cos 0 1 - 1 0)

Again, the indeterminate form (frac{0}{0}) is encountered. We apply L'H?pital's Rule a second time.

Second Application of L'H?pital's Rule

The numerator is:

(e^x - e^{sin x} cos x)

The derivative of (e^x) is (e^x) and the derivative of (e^{sin x} cos x) is (e^{sin x} cos x (cos x - sin x)). Therefore, the derivative of the numerator is:

(e^x - e^{sin x} (cos^2 x - sin x))

The denominator is:

(1 - cos x)

The derivative of the denominator is:

(sin x)

Status after second application:

(lim_{xto 0} frac{e^x - e^{sin x} (cos^2 x - sin x)}{sin x})

Again substituting (x 0), the numerator becomes:

(e^0 - e^{sin 0} (cos^2 0 - sin 0) 1 - 1(1 - 0) 0)

And the denominator becomes:

(sin 0 0)

The form is still indeterminate, requiring another application of L'H?pital's Rule. Given the complexity, let's further simplify using Taylor Series.

Taylor Series Approach

Using Taylor series expansions around (x 0):

(e^x approx 1 x frac{x^2}{2} frac{x^3}{6} O(x^4))

(sin x approx x - frac{x^3}{6} O(x^5))

The numerator becomes:

(1 x frac{x^2}{2} frac{x^3}{6} - left(1 left(x - frac{x^3}{6}right)(1 - x)right) frac{x^3}{6} O(x^4))

The denominator becomes:

(x - (x - frac{x^3}{6}) frac{x^3}{6} O(x^5))

Status after simplification:

(lim_{xto 0} frac{frac{x^3}{6} O(x^4)}{frac{x^3}{6} O(x^5)} 1)

Conclusion

The limit evaluates to 1, demonstrating the power of both L'H?pital's Rule and Taylor Series in resolving complex indeterminate forms. Understanding these techniques is crucial for advanced calculus and real-world applications in engineering and physics.