This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 137 WEEK 3 NOTES PROF. DOUG PARK 1. Limit laws Limit Laws . Suppose f ( x ) → L and g ( x ) → M , as x → a . Then • Sum Law: f ( x ) + g ( x ) → L + M • Difference Law: f ( x ) g ( x ) → L M • Constant Multiple Law: cf ( x ) → cL , where c = constant. • Product Law: f ( x ) · g ( x ) → L · M • Quotient Law: f ( x ) g ( x ) → L M , as long as M 6 = 0. • Composition Law: g ( f ( x )) → g ( L ), as long as g ( x ) is continuous at x = L . We will prove the Sum Law using the precise definition of limit. Proofs of other laws are found in Appendix F of Textbook. Proof of Sum Law . We need to prove lim x → a ( f ( x ) + g ( x )) = L + M if lim x → a f ( x ) = L and lim x → a g ( x ) = M . Given any 1 > 0, there is δ 1 > 0 such that <  x a  < δ 1 = ⇒  f ( x ) L  < 1 Given any 2 > 0, there is δ 2 > 0 such that <  x a  < δ 2 = ⇒  g ( x ) M  < 2 Given any > 0, let δ = min { δ 1 ,δ 2 } . Then <  x a  < δ = ⇒  f ( x ) L  < 1 and  g ( x ) M  < 2 = ⇒  f ( x ) L  +  g ( x ) M  < 1 + 2 = ⇒  ( f ( x ) L ) + ( g ( x ) M )  ≤  f ( x ) L  +  g ( x ) M  < 1 + 2 = ⇒  ( f ( x ) + g ( x )) ( L + M )  < 1 + 2 We are done if 1 + 2 = . Thus we set 1 = 2 = 2 . 2. Computing limits of form Ex . Evaluate L = lim x → a x 3 a 3 x a ( a = constant). Sol’n . If we plug in x = a , we get form. In such cases we try to cancel the common zero factor. x 3 a 3 = ( x a )( x 2 + ax + a 2 ) Date : September 28, 2009. 1 2 DOUG PARK Thus L = lim x → a x a x a · ( x 2 + ax + a 2 ) = lim x → a ( x 2 + ax + a 2 ) = a 2 + a 2 + a 2 = 3 a 2 Ex . Evaluate L = lim x → 2 x √ 3 x 2 √ x + 2 2 . Sol’n . form so we cancel ( x 2) terms. The ( x 2) terms are found when we multiply by conjugate expressions! Hence multiply both top and bottom by ( x + √ 3 x 2)( √ x + 2 + 2). We get ( x 2 3 x + 2)( √ x + 2 + 2) ( x + 2 4)( x + √ 3 x 2) = ( x 1)( x 2)( √ x + 2 + 2) ( x 2)( x + √ 3 x 2) = ( x 1)( √ x + 2 + 2) x + √ 3 x 2→ (2 1)( √ 2 + 2 + 2) 2 + √ 6 2 = 4 4 = 1 3. Onesided limits Onesided limits will usually be of constant form, or limits involving   symbol. Theorem . lim x → a f ( x ) = L ⇐⇒ lim x → a f ( x ) = lim x → a + f ( x ) = L Ex . Evaluate the following limits if they exist....
View
Full
Document
This note was uploaded on 12/18/2010 for the course ECONOMICS 120 taught by Professor Mesta during the Spring '10 term at Wilfred Laurier University .
 Spring '10
 Mesta

Click to edit the document details