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1.2 Finding Limits Graphically and Numerically

Calculus. 1.2 Finding Limits Graphically and Numerically. Numerical Investigation. Create a table for x values approaching 0. Graphic Investigation. Find the limit of f(x) as x approaches 2 where f is defined as. Non-existance. Example 3:.

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1.2 Finding Limits Graphically and Numerically

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  1. Calculus 1.2 Finding Limits Graphically and Numerically

  2. Numerical Investigation Create a table for x values approaching 0

  3. Graphic Investigation Find the limit of f(x) as x approaches 2 where f is defined as

  4. Non-existance Example 3: Solution: Consider the graph of the function f(x) = |x|/x. |x|/x =1, x>0 |x|/x =-1, x<0

  5. Non-existance Example 4: Discuss the existence of the limit: Solution: Using a graphical representation, you can see that x does not approach any number. Therefore, the limit

  6. Non-existance Example 5: Make a table approaching 0 The graph oscillates, so no limit exists.

  7. 1) f(x) approaches a different number from the right side of c than it approaches from the left side. Common types of behavior associated with nonexistence of a limit 2) f(x) increases or decreases without bound as x approaches c. 3) f(x) oscillates between two fixed values as x approaches c. Homework

  8. Definition of Limit Let f be a function defined on an open interval containing c (except possibly at c) and let L be a real number. The statement: lim f(x) = L X→c Means that for each e > 0 there exists a d >0 such that if 0 < |x-c| < d, then |f(x)-L|< e

  9. Means that for each e > 0 there exists a d >0 such that if 0 < |x-c| < d, then |f(x)-L|< e |f(x)-L| e |x-c| d Ex. 7 Ex. 8

  10. Thus, for a given e > 0 you can choose d = e/3. This choice works because Lim (3x-2) = 4 0< |x-2| < d = e/3 X→2 Example 7: Prove: Solution: You must show that for each e > 0, there exists a d>0 such that |(3x-2)-4)| < e. Because your choice of d depends on e, you need to establish a connection between |(3x-2)-4| and |x-2|. |(3x-2)-4| = 3|x-2| < 3(e/3) = e |(3x-2)-4| = |3x-6| = 3|x-2| Diagram

  11. Lim x2 = 4 Solution: You must show that for each e > 0, there exists a d > 0 such that Example 8: Use the e-d definition of a limit to prove that: X→2 when 0 < |x-2| < d, |x2 - 4| < e. So, 0 < |x-2| < d = e/5 HW 1.2 p. 53/3-40 (Just find the limits), 42-46 Diagram

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