Engineering Mathematics Lecture 3 Limits Fall Semester 2018
Delve into the world of engineering mathematics with Lecture 3 on Limits from the Fall Semester 2018. Instructor Twana A. Hussein, a skilled MSc in Structural Engineering, delivers valuable insights and knowledge essential for understanding the foundational concepts of limits in engineering mathematics.
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ENGINEERING MATHEMATICS Lecture 3 / Limits Fall Semester 2018 Instructor Instructor Twana Twana. A. Hussein . A. Hussein MSc . Structural Engineering MSc . Structural Engineering Twana.ahmad@ishik.edu.iq Twana.ahmad@ishik.edu.iq Civil Engineering Dept. Civil Engineering Dept. Ishik University Ishik University
Ishik University INTRODUCTIONTO LIMITS Definition: The limit of a function f(x) at some point x0 exists and is equal to L if and only if every small interval about the limit L, say the interval (L , L+ ), means you can find a small interval about x0, say the interval (x0 , x0 + ), which has all values of f(x) existing in the former small interval about the limit L, except possibly at x0 itself. 2 .
Ishik University INTRODUCTIONTO LIMITS Example-1: ???????? ? ? = 3? 2 Find the limit as x approaches 2. ( ( ) ) 2 x x lim 3 2 The limit does exist x -1 0 1 2 y -5 -2 1 4 The functions continues at x = 2 3 .
Ishik University INTRODUCTIONTO LIMITS Example 2: ???????? ?(?) = ?2 ? 6 Find the limit as x approaches 1. = = 2 x x l m i x 6 1 x -3 -2 -1 0 1 2 3 4 y 6 0 -4 -6 -6 -4 0 6 4 Fact: the limit does exist for all polynomials. .
Ishik University INTRODUCTIONTO LIMITS Example 3: ???????? 2 x x 6 ( ( ) ) f x = = x 3 2 x x 6 replace x with 1 Find = = li x m x 3 1 2 (1) (1) 6 = = = = lim x 3 (1) 3 1 5 .
Ishik University INTRODUCTIONTO LIMITS Example 3: ???????? 2 x x 6 ( ( ) ) f x = = x 3 Now, what happen if we find limit as x approaches 3? 2 (3) (3) 6 (3) 3 0 0 Now 0/0 is a difficulty! We don't really know the value of 0/0 "indeterminate"), need another answering this. = = = = li x m 3 (it is we of so way 6 .
Ishik University INTRODUCTIONTO LIMITS Though f(x) is not defined at x = 3, arbitrarily close to 3, ?(?) = ? + 2 . So as x approaches 3. + + 2 x x x x 6 ( 3 x )( 2) = = = = = = = = = = + + x lim x lim x lim x 2 x 3 3 3 3 3 ? ? = ? + ? = ? + ? = ? This function goes to 5. Its limit exists, but the function is not defined at x = 3. 7 .
Ishik University INTRODUCTIONTO LIMITS x y -1 1 0 2 1 3 2 4 3 5 8 .
Ishik University INTRODUCTIONTO LIMITS Example 4: ???????? + + 2 x x 12 = = ( ) f x x 3 What happens at x = 3? ??+ ? ?? ? ? =? ? ? = ? The denominator = 0 therefore, the functions is undefined. 9 .
Ishik University INTRODUCTIONTO LIMITS x 2 y 6 x y As for ? = 2.9 ? = 6.9 And for ? = 3.1 ? = 7.1 This circle means that there is a hole at point (3,7). 2.5 2.6 2.7 2.8 2.9 6.5 6.6 6.7 6.8 6.9 2.2 2.4 2.6 2.8 6.2 6.4 6.6 6.8 This tell us that for ? = ? ? = ? Undefine d 7.2 7.4 7.6 7.8 8 Undefine d 7.1 7.2 7.3 7.4 7.5 3 3 ??+? ?? ? ? ? ? ?+? ? ? ?(?) = 3.1 3.2 3.3 3.4 3.5 3.2 3.4 3.6 3.8 4 ?(?) = ? ? = ? + ? ? ? = ? + ? = ? 10 .
Ishik University INTRODUCTIONTO LIMITS Example 5: Consider 2 x x 2 7 15 = = ( ) f x x 5 What happens at x = 5? ? ?? ?(?) ?? (?) ? =? ? ? = ? The denominator = 0 therefore, the functions is undefined. 11 .
Ishik University INTRODUCTIONTO LIMITS + + x x ( 5)(2 x 3) = = = = + + x lim x lim(2 x 3) 5 5 5 + + = = lim(2(5) x 3) 13 5 The limit does exist but there is a hole at point (5,13) 12 .
Ishik University INTRODUCTIONTO LIMITS Example-6: ???????? As x = 9, we will have 0 x 9 = = lim x x 3 9 0case + + + + x x x 9 3 3 = = l m i x x 3 9 ( ( x ) ) + + x x 9 3 = = = = = = + + x lim x lim x 3 ( 9 ) 9 9 The limit exists for this function. However there s a hole at point (9, 6). = = + + = = lim x 9 3 6 9 13 .
