Evaluating Limits Involving Integrals: A Comprehensive Guide

When dealing with limits that involve integrals, L'hopital's Rule and the Fundamental Theorem of Calculus (FTC) are powerful tools that can simplify the process. Consider the problem of evaluating the limit:

lim_{x rightarrow 0} frac{int_{0}^{x^2} arctan t , dt}{int_{0}^{x} t^2 sin t , dt}

Applying L'hopital's Rule

For this type of limit, which is in the form of 0/0, we can apply L'hopital's Rule. The steps are as follows:

Take the derivative of the numerator and the denominator separately: Numerator: The derivative of int_{0}^{x^2} arctan t , dt is given by the FTC: [ frac{d}{dx} left( int_{0}^{x^2} arctan t , dt right) 2x arctan x^2 ] Denominator: The derivative of int_{0}^{x} t^2 sin t , dt is: [ frac{d}{dx} left( int_{0}^{x} t^2 sin t , dt right) x^2 sin x ] Thus, applying L'hopital's Rule, we get: [ lim_{x rightarrow 0} frac{2x arctan x^2}{x^2 sin x} ] Now, considering the limits of individual terms as x rightarrow 0: [ lim_{x rightarrow 0} 2x 0 [ lim_{x rightarrow 0} arctan x^2 frac{pi}{2} cdot x^2 approx x^2 (for x rightarrow 0) [ lim_{x rightarrow 0} x^2 sin x 0 (using lim_{x rightarrow 0} frac{sin x}{x} 1) When we combine these, we get: [ lim_{x rightarrow 0} frac{2x arctan x^2}{x^2 sin x} 2 cdot frac{frac{pi}{2} cdot x^2}{x^2} cdot frac{x}{x} 2 cdot 1 cdot 1 2 ]

Alternative Solution: Integration by Parts

Another approach is to use integration by parts, which is particularly useful for nested integrals. Consider the following steps:

Apply integration by parts to the integrals separately. For the numerator: ( int_{0}^{x^2} arctan t , dt ), let ( u arctan t ) and ( dv dt ). Then ( du frac{1}{1 t^2} , dt ) and ( v t ). For the denominator: ( int_{0}^{x} t^2 sin t , dt ), use integration by parts multiple times or consider a Maclaurin series expansion. Utilize the Maclaurin series expansion for each part. Combine the results and simplify, which eventually yields: [ frac{x^2 arctan x^2 - frac{1}{2} ln(1 x^4)}{(2-x^2) cos x^2 sin x^2 - 2} ] Take the limit as ( x rightarrow 0:) [ lim_{x rightarrow 0} frac{x^2 arctan x^2 - frac{1}{2} ln(1 x^4)}{(2-x^2) cos x^2 sin x^2 - 2} frac{0 - 0}{2 - 0 - 2} frac{0}{0} ] Which simplifies to: [ frac{3}{2} ]

Fundamental Theorem of Calculus Application

The Fundamental Theorem of Calculus (FTC) is essential in transforming integrals into functions whose derivatives are known. For instance:

( F(x) int_{0}^{x} arctan t , dt ) , then ( F'(x) arctan x ) ( G(x) int_{0}^{x} t^2 sin t , dt ) , then ( G'(x) x^2 sin x ) Using these, we can apply L'hopital's Rule effectively as explained in the previous solution.