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Week 3 Course Notes

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

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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 0 < | x - a | < δ 1 = | f ( x ) - L | < 1 Given any 2 > 0, there is δ 2 > 0 such that 0 < | x - a | < δ 2 = | g ( x ) - M | < 2 Given any > 0, let δ = min { δ 1 , δ 2 } . Then 0 < | 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 0 0 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 0 0 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

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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 . 0 0 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 0 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. (i) lim x 1 1 ( x - 1) 3 (ii) lim x 1 1 | x - 1 | (iii) lim x 1 x - 1 | x - 1 | Strategy . Note that 1 0 + = + , 1 0 - = -∞ . Sol’n . (i) lim x 1 + 1 ( x - 1) 3 = 1 (1 + - 1) 3 = 1 (0 + ) 3 = 1 0 + = + lim x 1 - 1 ( x - 1) 3 = 1 (1 - - 1) 3 = 1 (0 - ) 3 = 1 0 - = -∞ Hence the limit does not exist (DNE). (ii) lim x 1 + 1 | x - 1 | = 1 | 0 + | = 1 0 + = + lim x 1 - 1 | x - 1 | = 1 | 0 - | = 1 0 + = + Hence the limit is + , but DNE as a real number.
MATH 137 3 (iii) If x 1 + , then x > 1, so x - 1 > 0. Thus x - 1 | x - 1 | = x - 1 x - 1 = 1 lim x 1 + x - 1 | x - 1 | = lim x 1 + 1 = 1 If x 1 - , then x < 1, so x - 1 < 0. Thus x - 1 | x - 1 | = x - 1 - ( x - 1) = - 1 lim x 1 - x - 1 | x - 1 | = lim x

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