limit c test - ↓ = lim x →∞ 6 x 1 / 4 e x/ 2 L’H...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 101 – Fall 2010 The Improved Limit Comparison Test for Improper Integrals Section 8.7 Theorem 3* – The Limit Comparison Test: Let f and g be nonnegative functions on [ a, ) , and suppose L = lim x →∞ f ( x ) g ( x ) . Then: a. If 0 < L < , then: a f ( x ) dx converges ⇐⇒ a g ( x ) dx converges a f ( x ) dx diverges ⇐⇒ a g ( x ) dx diverges b. If L = 0 , then: a g ( x ) dx converges = a f ( x ) dx converges a f ( x ) dx diverges = a g ( x ) dx diverges c. If L = , then: a f ( x ) dx converges = a g ( x ) dx converges a g ( x ) dx diverges = a f ( x ) dx diverges Remark: Similar tests are valid for the other types of basic improper integrals. Examples: Determine whether each of the following improper integrals converges or diverges. 1. 0 x 3 e - x dx : We will do limit comparison with 0 e - x/ 2 dx . L = lim x →∞ x 3 e - x e - x/ 2 = lim x →∞ x 3 e x/ 2 L’H = lim x →∞ 3 x 2 1 / 2 e x/ 2 L’H
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ↓ = lim x →∞ 6 x 1 / 4 e x/ 2 L’H ↓ = lim x →∞ 6 1 / 8 e x/ 2 = 0 . We also have ∫ ∞ e-x/ 2 dx = lim c →∞ ∫ c e-x/ 2 dx = lim c →∞ [-2 e-x/ 2 ] c = lim c →∞ (-2 e-c/ 2 + 2) = 2 . Since L = 0 and ∫ ∞ e-x/ 2 dx converges, we conclude that ∫ ∞ x 3 e-x dx converges by the Limit Comparison Test. 2. ∫ ∞ π sin 2 x x 2 dx : We have L = lim x →∞ sin 2 x/x 2 1 /x 3 / 2 = lim x →∞ sin 2 x x 1 / 2 = 0 by the Sandwich Theorem. We also know that the p-integral ∫ ∞ π dx x 3 / 2 converges as p = 3 / 2 > 1 . Therefore ∫ ∞ π sin 2 x x 2 dx converges by the Limit Comparison Test....
View Full Document

This note was uploaded on 12/29/2010 for the course MATH 101 taught by Professor Okantekman during the Spring '10 term at Bilkent University.

Ask a homework question - tutors are online