Week 3 Course Notes - MATH 137 WEEK 3 NOTES PROF. DOUG PARK...

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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. One-sided limits One-sided 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....
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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 .

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Week 3 Course Notes - MATH 137 WEEK 3 NOTES PROF. DOUG PARK...

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