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Understanding Limits: Evaluating Function Behavior as Inputs Approach Specific Values

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milos
30 diapositives · 470 Views · Jul 30, 2014

This guide explores the fundamental concept of limits in calculus, describing how to determine the behavior of a function's outputs (y-values) as its inputs (x-values) approach a particular point. We'll cover various methods of finding limits, including direct substitution, factoring, and rationalizing. Continuous functions behave predictably, while discontinuities reveal the function's undefined points or asymptotic behavior. Examples will illustrate these principles and techniques to help clarify the understanding of limits within polynomial functions and rational expressions.

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Understanding Limits: Evaluating Function Behavior as Inputs Approach Specific Values
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  1. Determining Limit Values • Limits give us a language for describing how the outputs (y values) of a function behave as the inputs (xvalues) approachsome particular value. • Sometimes we use direct substitution, factoring, rationalizing or expanding to calculate a limit.

  2. Definition of a Limit We write limf(x) = L x→a and say “the limit of f(x), as xapproachesa, equals L” if we can make the values off(x) arbitrarily close to L (as close to Las we like) by taking xto be sufficiently close to a, but not equal to a.

  3. Functions Continuous at ‘a’ A function is continuous when its graph is a single unbroken curve that you could draw without lifting your pen from the paper. That is not a formal definition, but it helps you understand the idea.

  4. Functions Continuous at ‘a’ Polynomial Functions are Continuousat all x values (1, 5) (1 3) 5 3 lim5 = x→1 5 lim (– 2x + 5) = x→3 3

  5. Functions Continuous at ‘a’ lim() = x→3

  6. Functions Continuous at ‘a’ 0 0 limf(x) = x→2 (2, 0)

  7. If direct substitution can be done without an error we can determine the function value and limit this way. lim(7 – 2x) = x→3 7 – 2(3) = 1 7 – 2(3) = 1 42 + 2(4) = 24 lim(x2 + 2x) = x→4 42 + 2(4) = 24 lim(2x2 – 3x + 4) = x→2 2(2)2– 3(2) + 4 = 6 2(2)2– 3(2) + 4 = 6

  8. Limits Rational Expressions If ris a rational function given by and c is a real number such that q(c) ≠ 0, then

  9. Let’s look at some examples to clarify this MATH LANGUAGE

  10. Remember: Division by 0 is undefined. Rational Functions are continuous at any x value that does not result in division by zero.

  11. Try it! By Substitution The function is continuous at x = 3

  12. So what is not continuous (also called discontinuous)? Look out for holes, jumps or vertical asymptotes.

  13. Exploring a Nonexistent Limit. Rational Functions are discontinuous at any x value that DOES result in division by zero. undefined There will be a vertical asymptote when both function and limit do not exist

  14. Exploring a Nonexistent Limit. x = 1 Use a table to show the limit does not exist. Find • Use a graph to show the • limit does not exist. The y values are getting further apart the closer x gets to 1 Since f (1) does not exist and the limit as x→1 does not exist there is a vertical asymptote at x = 1

  15. Finding Limits Algebraically when the Function is Undefined Direct substitution won’t WORK!!! DNE Factor and reduce There is a vertical asymptote f(5) does not exist Substitute – 5 in forx Undefined!

  16. Finding Limits Algebraically when the Function is Undefined Direct substitution won’t WORK!!! undefined HOWEVER as x→5

  17. Finding Limits Algebraically when the Function is Undefined Factor and reduce f(5) does not exist There is a “hole” at Substitute 5 in for x

  18. Lesson #5 WorksheetEXAMPLE 1: The function DOES NOT exist at ‘a’. -∞ Approaching 0.1666.. ∞

  19. Lesson #5 Worksheet EXAMPLE 1: Continued The function f(x) does not exist at x= +3 but the limits of f(x) as x approaches are a different matter. (-3, 0.166666) x = 3 undefined

  20. Limit of a Piecewise Function The function g has a limit of 2 as x →1 even though g(1) ≠ 2 (1, 2) (1, 1) (1, 1) The function is discontinuous when x = 1. There is a “hole” at (1, 2).

  21. YES! Limit of a Function Involving a Radical Is there a radical rule? Let n be a positive integer. The following limit is valid for all c if n is odd, and is valid for c > 0 if n is even.

  22. n is odd and > 0 .44

  23. Finding Limits By Graphing By Substitution (3, 2) 2 3

  24. Use a graph of to show that the following limit does not exist. Exploring a Nonexistent Limits There is no point on the graph wherex= – 3 By substitution: is a non-real number

  25. Remember This! Multiply by its conjugate 3 NOTE: The answer is a rational number

  26. Rationalizing The Numerator Rational number

  27. Find the following Limit by Rationalizing The Numerator Direct substitution won’t WORK!!! 1

  28. Lesson 5 continued—Example 3 Direct substitution won’t WORK!!! Multiply Top and Bottom by the conjugate. 1

  29. Lesson 5 Worksheet continued Example 2 Page 2 Practice Rationalizing: 1

  30. EXAMPLE 3: Expanding Technique Again substitution won’t WORK. Find