Proving the Limit Involving the Integral of a Monotonically Decreasing Function

Understanding the behavior of mathematical functions and their integrals is crucial in many advanced mathematical applications. In this article, we delve into the intriguing problem of proving the following limit:

( limlimits_{h to 0} frac{1}{h} intlimits_{x}^{xh} frac{1}{t} , dt frac{1}{x} ) for ( x > 0 )

Understanding Monotone Functions and Their Integrals

A function ( f(t) frac{1}{t} ) is a monotonically decreasing function for ( t > 0 ). This means that as ( t ) increases, ( f(t) ) decreases. Therefore, we can deduce that for any ( t ) between ( x ) and ( xh ) where ( h > 0 ), the inequality

( frac{1}{x} geq frac{1}{t} geq frac{1}{xh} )

holds. This implies that the integral of ( frac{1}{t} ) over the interval ( [x, xh] ) must lie between the integrals of ( frac{1}{xh} ) and ( frac{1}{x} ) over this interval. As ( h ) approaches 0, these bounds also approach 0, which explains why the original statement is false.

Proving the Limit Using the Fundamental Theorem of Calculus

Let's prove the limit using the Fundamental Theorem of Calculus (FTC), which states that if ( f ) is a continuous function over some interval ( I ), then the function ( F(x) ) defined as ( F(x) intlimits_{a}^{x} f(t) , dt ) for fixed ( a in I ) is differentiable with ( F'(x) f(x) ).

Define ( I(x) intlimits_{x}^{xh} frac{1}{t} , dt ). By the FTC, the derivative of ( I(x) ) is:

( I'(x) frac{d}{dx} left( intlimits_{x}^{xh} frac{1}{t} , dt right) frac{1}{xh} cdot h - frac{1}{x} frac{1}{x} cdot (h - 1) frac{1}{x} )

For small ( h ), ( frac{I(x)}{h} ) approximates ( I'(x) ) as ( h ) approaches 0. Therefore:

( limlimits_{h to 0} frac{1}{h} intlimits_{x}^{xh} frac{1}{t} , dt limlimits_{h to 0} I'(x) frac{1}{x} )

Mean Value Theorem of Integrals

The Mean Value Theorem of Integrals states that if ( f(t) ) is continuous on ( [a, b] ), then there exists a point ( c ) in the interval ( [a, b] ) such that:

( frac{1}{b - a} intlimits_{a}^{b} f(t) , dt f(c) )

In our problem, ( a x ), ( b xh ), and ( f(t) frac{1}{t} ). The interval ( [x, xh] ) shrinks to ( [x, x] ) as ( h ) approaches 0. Therefore, ( c ) approaches ( x ), and we conclude:

( limlimits_{h to 0} f(c) frac{1}{x} )

Using Definition of Derivative

To further reinforce the proof, let's use the definition of the derivative:

( frac{1}{h} left( intlimits_{x}^{xh} frac{1}{t} , dt - intlimits_{x}^{x} frac{1}{t} , dt right) frac{1}{xh} left( intlimits_{x}^{xh} frac{1}{t} , dt - intlimits_{x}^{x} frac{1}{t} , dt right) )

By the FTC, this becomes:

( frac{1}{x} left( intlimits_{x}^{xh} frac{1}{t} , dt right) frac{1}{xh} cdot frac{1}{xh} - frac{1}{x} )

As ( h ) approaches 0, the expression simplifies to:

( limlimits_{h to 0} frac{1}{h} intlimits_{x}^{xh} frac{1}{t} , dt frac{1}{x} )

Note: Applying L'Hopital's Rule would be circular since differentiating the integral involves the fundamental theorem of calculus, which we are using to prove this very limit.

Conclusion

The key to understanding this problem lies in the fundamental theorem of calculus, the mean value theorem of integrals, and the definition of the derivative. By applying these concepts, we have rigorously proven the limit:

( limlimits_{h to 0} frac{1}{h} intlimits_{x}^{xh} frac{1}{t} , dt frac{1}{x} ) for ( x > 0 ).

Understanding such limits is crucial in advanced mathematical analysis, particularly in Calculus and Real Analysis courses.