8031137-Mathematics-Techniques-Topic-3-Limits-Continuity-Differentiation-T1-200809

# 8031137-Mathematics-Techniques-Topic-3-Limits-Continuity-Differentiation-T1-200809

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TMT1171 Mathematical Techniques I, Trimester 1 2008/2009 TOPIC 3: Limits, Continuity and Differentiation A. LIMITS OF A FUNCTION 1. LIMIT DEFINITION Definition: For a function ) ( x f f = defined on an open interval ( , b ) about a number c , except possibly at c itself, if gets arbitrarily close to a number L for all a ) x ( f x sufficiently close to c (on either side of c ) but not equal to c , then we say that approaches the limit as f L x approaches , and we write c L x f c x = ) ( lim or as L x f ) ( c x . and say “the limit of , as ) ( x f x approaches , equals ”. c L Example : Find the limit of 3x 2 -1 as x approaches 0. x f(x) x f(x) -0.1 -0.97 0.1 -0.97 -0.01 -0.9997 0.01 -0.9997 -0.001 -0.999997 0.001 -0.999997 -0.0001 -0.99999997 0.0001 -0.99999997 As x 0, f(x) -1. So, ( ) 1 1 3 lim 2 0 = x x If no such number L exists, we say that has no limit at c () does not exist). Notice that the limit does not depend on how the function is defined at c . The limit may exist even if the value of is not known or undefined at . f ( lim x f c x f c Example: Find the limit of and , as x approaches 2. = = 2 , 2 2 , ) ( 2 x x x x g > = 2 , 3 2 , ) ( 2 x x x x x h Solution: 4 ) ( lim 2 = x g x lim does not exist ) ( 2 x h x 1

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TMT1171 Mathematical Techniques I, Trimester 1 2008/2009 Definition: More formally, we say that the limit of as ) ( x f x approaches is if for every number c L 0 > ε there is a corresponding number 0 > = δ such that < L x f ) ( whenever < c x 0 LAWS < 2. LIMIT Suppose L x f c x = ) ( lim and x M x g c = ) ( m li . iqueness: implies L K = 1. Un f K x c x = ) ( lim , i.e. a function has at most one lim . Sum Rule: it at a particular number [ ] x g x f x g x f M L c x c x c x + = 2 + = + ) ( lim ) ( lim ) ( ) ( lim [ ] M L x g x f x g x f c x c x c x = = ) ( lim ) ( lim ) ( ) ( lim 3. Difference R le: u 4. Product Rule: M L x g x f x g x f c x c x c x = = ) ( lim ) ( lim ) ( ) ( lim 5. Constant Multi le Ru e L k x f k x kf c x c x = = ) ( lim ) ( lim R k p l : for any 6. Quotient Rule: M L x g x f x g x f c x c x c x = = ) ( lim ) ( lim ) ( ) ( lim provided 0 M M x g c x 1 ) ( 1 lim = provided M 0 7. Power Rule: ( ) s r c x x f / ) ( lim = s r L / for any r Z s , with 0 s provided R L s r / 8. ,where is a positive integer a a c x = lim 9. c x c x = lim 10. n n c x c x = lim n 11. n n c x c x = ,where n is a positive intege lim r (and if is even, we assume that c > 0) n 12. n n c x lim L x f = ) ( ,where is a positive integer (and if is even, we assume that L > 0) n n xam E ple: Evaluate the following limits, if they exist. 2 2 a) ( ) 1 4 lim 2 2 + x x x b) lim 3 + x x x c) 4 4 lim 2 2 2 + x x x d) x x x x 3 5 1 2 lim 2 3 2 + 1 1 lim 2 1 x x x 1 4 1 2 lim 2 1 + x x x e) f) h) x x x 1 1 lim 0 + i) 3 4 lim 2 2 x x g) x x x 16 ) 4 ( lim 2 0 + k) ( ) 3 / 1 2 1 2 lim x x 3 2 lim 2 2 x x j) 2
TMT1171 Mathematical Techniques I, Trimester 1 2008/2009 Solution: ( ) 1 4 2 + x x a) lim 2 x ( ) 3 1 8 4 1 lim 4 lim lim 1 4 lim 2 2 2 2 2 2 = + = + + x x x x x x x x = 1 1 lim 2 1 x x x e) 2 1 ) 1 ( lim 1 1 1 lim ) 1 )( 1 ( ) 1 ( lim 1 1 lim 1 1 1 2 1 = + = + = + = x x x x x x x x x x x

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8031137-Mathematics-Techniques-Topic-3-Limits-Continuity-Differentiation-T1-200809

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