L4 one sided limits limits at infinity
- 2. LIMITS OF FUNCTIONS OBJECTIVES: •define limits; •illustrate limits and its theorems; and •evaluate limits applying the given theorems. • define one-sided limits • illustrate one-sided limits • investigate the limit if it exist or not using the concept of one-sided limits. •define limits at infinity; •illustrate the limits at infinity; and •determine the horizontal asymptote.
- 3. DEFINITION: LIMITS The most basic use of limits is to describe how a function behaves as the independent variable approaches a given value. For example let us examine the behavior of the function for x-values closer and closer to 2. It is evident from the graph and the table in the next slide that the values of f(x) get closer and closer to 3 as the values of x are selected closer and closer to 2 on either the left or right side of 2. We describe this by saying that the “limit of is 3 as x approaches 2 from either side,” we write 1xx)x(f 2 +−= 1xx)x(f 2 +−= ( ) 31xxlim 2 2x =+− →
- 4. 2 3 f(x) f(x) x y 1xxy 2 +−= x 1.9 1.95 1.99 1.995 1.999 2 2.001 2.005 2.01 2.05 2.1 F(x) 2.71 2.852 2.97 2.985 2.997 3.003 3.015 3.031 3.152 3.31 left side right side O
- 5. 1.1.1 (p. 70) Limits (An Informal View) This leads us to the following general idea.
- 6. EXAMPLE Use numerical evidence to make a conjecture about the value of . 1x 1x lim 1x − − → Although the function is undefined at x=1, this has no bearing on the limit. The table shows sample x-values approaching 1 from the left side and from the right side. In both cases the corresponding values of f(x) appear to get closer and closer to 2, and hence we conjecture that and is consistent with the graph of f. 1x 1x )x(f − − = 2 1x 1x lim 1x = − − →
- 7. Figure 1.1.9 (p. 71) x .99 .999 .9999 .99999 1 1.00001 1.0001 1.001 1.01 F(x) 1.9949 1.9995 1.99995 1.999995 2.000005 2.00005 2.0005 2.004915
- 8. THEOREMS ON LIMITS Our strategy for finding limits algebraically has two parts: •First we will obtain the limits of some simpler function •Then we will develop a list of theorems that will enable us to use the limits of simple functions as building blocks for finding limits of more complicated functions.
- 9. We start with the following basic theorems, which are illustrated in Fig 1.2.1 Theorem 1.2.1 (p. 80) ( ) ( ) axlimbkklima numbers.realbekandaLetTheorem1.2.1 axax == →→
- 10. Figure 1.2.1 (p. 80)
- 11. ( ) 33lim33lim33lim example,For a.ofvaluesallforaxaskf(x) whyexplainswhichvaries,xaskatfixedremain f(x)ofvaluesthethenfunction,constantaiskxfIf x0x-25x === →→ = →→→ π Example 1. ( ) ( ) π π =−== →→= →→→ xlim2xlim0xlim example,For .axfthattruebealsomustitaxthenx,xfIf x-2x0x Example 2.
- 12. Theorem 1.2.2 (p. 81) The following theorem will be our basic tool for finding limits algebraically.
- 13. This theorem can be stated informally as follows: a) The limit of a sum is the sum of the limits. b) The limit of a difference is the difference of the limits. c) The limits of a product is the product of the limits. d)The limits of a quotient is the quotient of the limits, provided the limit of the denominator is not zero. e) The limit of the nth root is the nth root of the limit. •A constant factor can be moved through a limit symbol.
- 14. ( )5x2lim.1 4x + → ( )12x6lim.2 3x − → ( ) )2x5(x4lim.3 3x −− → EXAMPLE : Evaluate the following limits. 31 58 5)4(2 5limxlim2 5limx2lim 4x4x 4x4x = += += += += →→ →→ ( ) 6 12-18 12)3(6 12limx6lim 3x3x = = −= −= →→ ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 13 131 2)3(534 2limxlim5xlim4lim 2limx5limxlim4lim 2x5limx4lim 3x3x3x3x 3x3x3x3x 3x3x = = −−= −⋅−= −⋅−= −•−= →→→→ →→→→ →→
- 15. 4x5 x2 lim.4 5x −→ ( )3 3x 6x3lim.5 + → 3x 1x8 lim.6 1x + + → ( ) 21 10 425 52 = − = ( ) ( ) ( ) ( )4limxlim5 xlim2 4limx5lim x2lim 5x5x 5x 5x5x 5x →→ → →→ → − = − = ( )( ) ( ) ( ) ( )( ) ( ) 3375 15633 6limxlim3 6limx3lim 6x3lim 33 3 3x3x 3 3x3x 3 3x = =+⋅= += += += →→ →→ → 2 3 4 9 3x 1x8 lim 1x == + + = →
- 16. OR When evaluating the limit of a function at a given value, simply replace the variable by the indicated limit then solve for the value of the function: ( ) ( ) ( ) 22 3 lim 3 4 1 3 3 4 3 1 27 12 1 38 x x x → + − = + − = + − =
- 17. EXAMPLE: Evaluate the following limits. 2x 8x lim.1 3 2x + + −→ Solution: ( ) 0 0 0 88 22 82 2x 8x lim 33 2x = +− = +− +− = + + −→ Equivalent function: (indeterminate) ( )( ) 2x 4x2x2x lim 2 2x + +−+ = −→ ( ) ( ) ( ) 12444 4222 4x2xlim 2 2 2x =++= +−−−= +−= −→ 12 2x 8x lim 3 2x = + + ∴ −→
- 18. Note: In evaluating a limit of a quotient which reduces to , simplify the fraction. Just remove the common factor in the numerator and denominator which makes the quotient . To do this use factoring or rationalizing the numerator or denominator, wherever the radical is. 0 0 0 0
- 19. x 22x lim.2 0x −+ → Solution: Rationalizing the numerator: (indeterminate) 0 0 0 220 x 22x lim 0x = −+ = −+ → ( )22xx 22x lim 22x 22x x 22x lim 0x0x ++ −+ = ++ ++ • −+ = →→ ( ) 4 2 22 1 22 1 22x 1 lim 22xx x lim 0x0x == + = ++ = ++ = →→ 4 2 x 22x lim 0x = −+ ∴ →
- 20. 9x4 27x8 lim.3 2 3 2 3 x − − → Solution: By Factoring: (indeterminate)3 2 3 3 22 3 8 27 8 27 27 27 02 lim 4 9 9 9 03 4 9 2 x x x→ − ÷− − = = = − − − ÷ ( )( ) ( )( ) ( ) + + + = + ++ = −+ ++− = →→ 3 2 3 2 9 2 3 6 2 3 4 3x2 9x6x4 lim 3x23x2 9x6x43x2 lim 2 2 2 3 x 2 2 3 x 2 23 2 3 2 9 6 27 33 999 ==== + ++ = 2 23 9x4 27x8 lim 2 3 2 3 x = − − ∴ →
- 21. 5x 3x2x lim.4 2 3 2x + ++ → Solution: ( ) ( ) ( ) 33 222 2 2 2 32 3 lim 5 2 5 8 4 3 4 5 15 9 15 3 x x x x→ + ++ + = + + + + = + = = 3 15 5x 3x2x lim 2 3 2x = + ++ ∴ →
- 22. DEFINITION: One-Sided Limits The limit of a function is called two-sided limit if it requires the values of f(x) to get closer and closer to a number as the values of x are taken from either side of x=a. However some functions exhibit different behaviors on the two sides of an x-value a in which case it is necessary to distinguish whether the values of x near a are on the left side or on the right side of a for purposes of investigating limiting behavior.
- 23. Consider the function <− > == 0x,1 0x,1 x x )x(f 1 -1 As x approaches 0 from the right, the values of f(x) approach a limit of 1, and similarly , as x approaches 0 from the left, the values of f(x) approach a limit of -1. 1 x x limand1 x x lim ,symbolsIn oxox −== −+ →→
- 24. 1.1.2 (p. 72) One-Sided Limits (An Informal View) This leads to the general idea of a one-sided limit
- 25. 1.1.3 (p. 73) The Relationship Between One-Sided and Two-Sided Limits
- 26. EXAMPLE: x x )x(f =1. Find if the two sided limits exist given 1 -1 exist.notdoes x x limor existnotdoesitlimsidedtwothethen x x lim x x limthecesin 1 x x limand1 x x lim ox oxox oxox → →→ →→ −+ −+ ≠ −== SOLUTION
- 27. EXAMPLE: 2. For the functions in Fig 1.1.13, find the one-sided limit and the two-sided limits at x=a if they exists. The functions in all three figures have the same one-sided limits as , since the functions are Identical, except at x=a. ax → 1)x(flimand3)x(flim areitslimThese axax == −+ →→ In all three cases the two-sided limit does not exist as because the one sided limits are not equal.ax → SOLUTION
- 28. Figure 1.1.13 (p. 73)
- 29. 3. Find if the two-sided limit exists and sketch the graph of 2 6+x if x < -2 ( ) = x if x -2 g x ≥ ( ) 4 26 x6lim)x(glim.a 2x2x = −= += −− −→−→ ( ) 4 2- xlim)x(glim.b 2 2 2x2x = = = ++ −→−→ 4)x(glimor 4toequalisandexistitlimsidedtwothethen )x(glim)x(glimthecesin 2x 2x2x = = −→ −→−→ −+ SOLUTION EXAMPLE:
- 30. x -2-6 4 y 4
- 31. 4. Find if the two-sided limit exists and sketch the graph of and sketch the graph. 2 2 3 + x if x < -2 ( ) = 0 if x = -2 11 - x if x > -2 f x SOLUTION ( ) ( ) 7 23 x3lim)x(flim.a 2 2 2x2x = −+= += −− −→−→ ( ) ( ) 7 2-11 x11lim)x(flim.b 2 2 2x2x = −= −= ++ −→−→ 7)x(flimor 7toequalisandexistitlimsidedtwothethen )x(flim)x(flimthecesin 2x 2x2x = = −→ −→−→ −+ EXAMPLE:
- 32. graph.thesketchand ,existf(x)limifeminerdet,4x23)x(fIf.5 2x→ −+= ( ) 3 4223 4x23lim)x(flim.a 2x2x = −+= −+= −− →→ ( ) 3 4223 4x23lim)x(flim.b 2x2x = −+= −+= ++ →→ 3)x(flimor 3toequalisandexistitlimsidedtwothethen )x(flim)x(flimthecesin 2x 2x2x = = → →→ −+ SOLUTION EXAMPLE:
- 33. f(x) x (2,3) 2
- 34. DEFINITION: LIMITS AT INFINITY The behavior of a function as x increases or decreases without bound is sometimes called the end behavior of the function. )x(f If the values of the variable x increase without bound, then we write , and if the values of x decrease without bound, then we write . +∞→x −∞→x For example , 0 x 1 limand0 x 1 lim xx == +∞→−∞→
- 36. 1.3.1 (p. 89) Limits at Infinity (An Informal View) In general, we will use the following notation.
- 37. Figure 1.3.2 (p. 89) Fig.1.3.2 illustrates the end behavior of the function f when L)x(flimorL)x(flim xx == −∞→+∞→
- 38. Figure 1.3.4 (p. 90) EXAMPLE Fig.1.3.2 illustrates the graph of . As suggested by this graph, x x 1 1y += e x 1 1lim ande x 1 1lim x x x x = + = + −∞→ +∞→
- 39. EXAMPLE ( Examples 7-11 from pages 92-95) 6x3 2x lim.4 x31 1x2x5 lim.3 5x2 xx4 lim.2 8x6 5x3 lim.1 2 x 23 x 3 2 x x − + − +− − − − + +∞→ +∞→ −∞→ +∞→ ( ) ( )336 x 36 x xx5xlim.6 x5xlim.5 −+ −+ +∞→ +∞→
- 40. EXERCISES: ( ) ( )( ) ( ) 5w4w 7w7w lim10. 2x 8x lim.5 19x9x2lim9. 4y y8y4 lim.4 1y2y 3y2y1y lim8. 1x 4x3x lim.3 1x 3x2x3x2 lim7. 4x3x 1x2 lim.2 1x9 1x3 lim6.2x5x4lim.1 2 2 1w 3 2x 2 1 34 5x 3 1 3 2y 2 2 1y3 2 1x 2 23 1x21x 2 3 1 x 2 3x −− ++ − − +− + + +− −+− + ++ − −−+ +− + − − +− −→→ − →→ →−→ →−→ → → A. Evaluate the following limits.
- 41. EXERCISES: B. Sketch the graph of the following functions and the indicated limit if it exists. find . )x(glim.cg(x)lim.bg(x)lim.a 1xif2x-7 1xif2 1xif3x2 )x(g.2 )x(flim.cf(x)lim.bf(x)lim.a 4-xif4x -4xifx4 )x(f.1 1x1x1x 4x4x4x →→→ −→−→−→ −+ −+ > = <+ = ≤+ >− =