Chapter The function f and its graph are shown below: + < x. lim f ( x) (a) Calculate. (b) Which value is greater

Transcription

1 Chapter The function f and its graph are shown below: f( ) = < 0 1 = 1 1< < 3 (a) Calculate lim f ( ) (b) Which value is greater lim f ( ) or f (1)? Justify your conclusion. (c) At what value(s) of c on the interval [ ] 0, 4 does lim f ( ) not eist? Justify your conclusion.

2 Chapter 1 1. The function f and its graph are shown below: f( ) = < 0 1 = 1 1< < 3 (a) Calculate lim f ( ) As we approach = from the left, the graph of f ( ) nears 4 so lim f( ) = 4. 1 lim f( ) = 4 (b) Which value is greater lim f ( ) or f (1)? Justify your conclusion. Since lim f( ) = 1 and lim f( ) = 1, lim f( ) = 1. From the function equation and graph, we see f (1) =. Therefore, f (1) > lim f( ). (c) At what value(s) of c on the interval [ ] 0, 4 does lim f ( ) not eist? Justify your conclusion. If the function f is continuous at = c, then lim f ( ) eists and lim f ( ) = f( c). We find the discontinuities in the graph of f ( ). Discontinuities occur at = 1 and =. Consider lim f ( ) and lim f ( ). In part (b), we showed lim f( ) = 1 so the limit eists at = 1. Consider lim f ( ). lim f( ) = 4 and lim f( ) = 3. Since the left and right hand limits are not equal, lim f ( ) does not eist. 1 lim f( ) = 1 1 f (1) = f (1) > lim f( ) 1 If f is continuous at = c, then lim f ( ) eists. 1 lim f( ) = 1 1 lim f( ) = 4 and lim f( ) = 3 1 At =, lim f ( ) does not eist.

3 Chapter 1 3. The function f and its graph are shown below: 0 < 1 3 = 1 f( ) = 1< < 1 (a) Calculate lim f ( ) (b) Which value is greater lim f ( ) or f ()? Justify your conclusion. (c) At what value(s) of c on the interval [ ] 0,3 does lim f ( ) not eist? Justify your conclusion.

4 Chapter 1 4. The function f and its graph are shown below: 0 < 1 3 = 1 f( ) = 1< < 1 (a) Calculate lim f ( ) As we approach = from the right, the graph of f ( ) nears 1 so lim f( ) = 1. 1 lim f( ) = 1 (b) Which value is greater lim f ( ) or f ()? Justify your conclusion. Since lim f( ) = and lim f( ) =, lim f( ) =. From the function equation and graph, we see f () = 1. Therefore, lim f ( ) > f(). 1 lim f( ) = 1 f () = 1 lim f ( ) > f() (c) At what value(s) of c on the interval [ ] Justify your conclusion. 0,3 does lim f ( ) not eist? If the function f is continuous at = c, then lim f ( ) eists and lim f ( ) = f( c). We find the discontinuities in the graph of f ( ). Discontinuities occur at = 1 and =. Consider lim f ( ) and lim f ( ). In part (b), we showed lim f( ) = so the limit eists at = 1. Consider lim f ( ). lim f( ) = 4 and lim f( ) = 1. Since the left and right hand limits are not equal, lim f ( ) does not eist. 1 If f is continuous at = c, then lim f ( ) eists. 1 lim f( ) = 1 lim f( ) = 4 and lim f( ) = 1 1 At =, lim f ( ) does not eist.

5 Chapter The function g is defined as follows: g ( ) = (a) Use a table of values to estimate lim g ( ) work that leads to your conclusion. accurate to three decimal places. Show the (b) Calculate lim g ( ) and lim g ( )? Show the work that leads to your conclusion. 3 3 (c) g ( ) is undefined at = and = 3. Eplain why not eist. lim g ( ) eists but lim g ( ) 3 does

6 Chapter The function g is defined as follows: g ( ) = (a) Use a table of values to estimate lim g ( ) accurate to three decimal places. Show the work that leads to your conclusion. g ( ) lim g ( ) = undef (b) Calculate lim g ( ) and lim g ( )? Show the work that leads to 3 your conclusion. 3 g ( ) lim g ( ) = ,001 lim g ( ) = 3 3 undef , (c) g ( ) is undefined at = and = 3. Eplain why eists but lim g ( ) does not eist. 3 lim g ( ) 4 The function g ( ) = may be rewritten as 5 6 ( )( ) g ( ) =. Observe that and 3 because ( )( 3) these values make the denominator zero. With this restriction, g ( ) may be further simplified to g ( ) =. This function has 3 a vertical asymptote at = 3 and a removable discontinuity at =. Therefore, lim g ( ) eists but lim g ( ) does not eist. 3 1 lim g ( ) = Table of values is used 1 Table of values shows two table entries on 1 1 either side of with g ( ) = lim g ( ) = 3 lim g ( ) = 3 Table of values or graph support conclusion 1 Eplains that g ( ) has a vertical asymptote at = 3. 1 Eplains that g ( ) has a removable discontinuity at =.

7 Chapter The function g is defined as follows: g ( ) = 4 (a) Use a table of values to estimate lim g ( ) work that leads to your conclusion. accurate to three decimal places. Show the (b) Calculate lim g ( ) and lim g ( )? Show the work that leads to your conclusion. (c) g ( ) is undefined at = and = not eist.. Eplain whylim g ( ) eists but lim g ( ) does

8 Chapter The function g is defined as follows: g ( ) =. 4 (a) Use a table of values to estimate lim g ( ) accurate to three decimal places. Show the work that leads to your conclusion. g ( ) lim g ( ) = undef (b) Calculate lim g ( ) and lim g ( )? Show the work that leads to your conclusion. g ( ).1 1 lim g ( ) = lim g ( ) = undef (c) g ( ) is undefined at = and =. Eplain whylim g ( ) eists but lim g ( ) does not eist. The function g ( ) = may be rewritten as 4 ( ) g ( ) =. Observe that and because ( )( ) these values make the denominator zero. With this restriction, g ( ) may be further simplified to g ( ) =. This function has a vertical asymptote at = and a removable discontinuity at =. Therefore, lim g ( ) eists but lim g ( ) does not eist. 1 lim g ( ) = Table of values is used 1 Table of values shows two table entries on 1 1 either side of with g ( ) = lim g ( ) = lim g ( ) = Table of values or graph support conclusion 1 Eplains that g ( ) has a vertical asymptote at =. 1 Eplains that g ( ) has a removable discontinuity at =.

9 Chapter Define 4 1 < 1 f( ) = < (a) Show f ( ) is continuous at = 5. (b) Where on the interval [ 1, 4 ] is f discontinuous? Show the work that leads to your conclusion. (c) Eplain the difference between a removable and an irremovable discontinuity.

10 Chapter Define 4 1 < 1 f( ) = < (a) Show f ( ) is continuous at = lim f( ) = lim f( ) = 0.5 so 5 5 to 0.5. f (5) = lim f ( ) = f(5) 5 lim f ( ) 5 eists and is equal 1 Show lim f ( ) eists 5 1 f (5) = lim f ( ) = f(5) 5 (b) Where on the interval [ 1, 4 ] is f discontinuous? Show the work that leads to your conclusion. Since linear functions are continuous, f is continuous on [1, ). Since lim f( ) = and lim f( ) = 1, lim f ( ) does not eist. Therefore, f is discontinuous at =. The 1 function is defined for all values of on [, 4] ecept at 3 = 3. Therefore, a discontinuity occurs at = 3. For every other value c in the interval [,4], lim f ( ) = f( c). Thus f is continuous at every other point. (c) Eplain the difference between a removable and an irremovable discontinuity. If lim f ( ) eists but lim f ( ) f( c), then a removable discontinuity occurs at = c. If lim f ( ) does not eist then an irremovable discontinuity occurs at = c. 1 Show that lim f ( ) does not eist. 1 States f is discontinuous at = 1 State f is discontinuous at = 3 because f is undefined there 1 Provide justification that f is continuous everywhere else. 1 Concept of lim f ( ) eists but lim f ( ) f( c) for removable 1 Concept of lim f ( ) does not eist for irremovable

11 Chapter Define ln 1 f( ) =. (a) Show f ( ) is continuous at =. (b) Where on the interval [, ] is f discontinuous? Show the work that leads to your conclusion. (c) Classify the discontinuities in part (b) as removable or irremovable.

12 Chapter Define ln 1 f( ) =. (a) Show f ( ) is continuous at = lim f( ) = lim f( ) = 0 so ln 1 ln 1 f () = = = 0 lim f ( ) = f() lim f ( ) eists and is equal to 0. 1 Show lim f ( ) eists 1 f () = 0 1 lim f ( ) = f() (b) Where on the interval [, ] is f discontinuous? Show the work that leads to your conclusion. We know y = ln is continuous for positive values of. Consequently, y = ln is continuous for all nonzero values of ; however, it is undefined and thus discontinuous when = 0. Similarly, y = ln 1 is continuous for all values of ecept at = 1, where it is undefined. Therefore, the function ln 1 f( ) = is discontinuous at = 1. It is also discontinuous at = 0 because f is undefined there. Shows that f is undefined at = 0 and = 1. 1 States that f is discontinuous where it is not defined. 1 Provides justification that f is continuous everywhere else. (c) Classify the discontinuities in part (b) as removable or irremovable. f ( ) f ( ) undef 1 undef In the table on the left, we see that lim f( ) = 1. Since the limit 0 eists, = 0 is a removable discontinuity. In the table on the right, it initially appears as if lim f( ) = 9.09 ; however, we notice that the values of f are changing rather dramatically as we near = 1. In fact, f ( ) = which is not at all close to Consequently, we conclude that lim f( ) =. Therefore, = 1 is an irremovable discontinuity. 1 = 0 is a removable discontinuity 1 = 1 is an irremovable discontinuity

13 Chapter Define f( ) = 1. 1 (a) Find lim f ( ) analytically. (b) Define 1 1 g ( ) =. What is the relationship between f and g? 1 (c) Eplain why lim f( ) =.

14 Chapter Define f( ) =. 1 (a) Find lim f ( ) analytically. 1 lim f( ) = lim 1 1 = lim 1 = lim ( 1) ( 1)( 1) ( 1) ( 1) 1 1 = 1 1 = 0 1 (b) Define g ( ) =. What is the relationship between f and g? 1 The graphs of f and g will look essentially the same. However, the graph of f will have a removable discontinuity at =1. or The domain of f is all real numbers ecept =± 1 whereas the domain of g is all real numbers ecept = 1. (c) Eplain why lim f( ) =. We evaluate f ( ) at values of increasingly close to = 1 as we approach = 1 from the right. f (1.1) = 1 f (1.01) = 01 f (1.001) = 001 f (1.0001) = 0001 We see that the values of f become increasingly large as nears = 1 from the right. Thus lim f( ) = 1 Correctly factor numerator and denominator. 1 Cancel 1 lim f( ) = 0 1 Observe that the graphs of f and g differ by a single point OR that the domains of f and g differ by a single value 1 Evaluate f at four or more values of within [ 1,1.1 ] Show a trend of increasing function values as nears 1 from the right.

15 Chapter Define cos 1 f( ) =. sin (a) Use a table of values to estimate lim f ( ) accurate to three decimal places. (b) Define g ( ) = ( ) 0 1 f. Use properties of limits to show that 1 lim g ( ) =. 0 (c) Is g continuous? Show the work that leads to your conclusion.

16 Chapter 1 16 cos 1 8. Define f( ) =. sin (a) Use a table of values to estimate three decimal places. lim f ( ) 0 accurate to f ( ) undef lim f( ) = (b) Define g ( ) =. Use properties of limits to f ( ) 1 show that lim g ( ) =. 0 In part (a), we showed that that lim ( 1) = 1. So 0 lim ( 1) 0 lim g ( ) = 0 lim f( ) 0 lim f( ) = 0. We also know 0 1 = 0 1 = (c) Is g continuous? Show the work that leads to your conclusion. 1 g ( ) = f ( ) 1 = cos 1 sin At = 0, sin = 0. This makes the denominator undefined. Therefore, g is not continuous at = 0 1 lim f( ) = Table of values is used 1 Table of values shows two table entries on either side of 0 with f( ) = ( ) lim 1 = lim f( ) = 0 1 properly apply limit properties to show 1 lim g ( ) = 0 1 Find a discontinuity State that function is not continuous.

17 Chapter Define 1 cos f( ) = and g ( ) = on the interval (0, ). (a) Show g( ) f ( ) 0 (b) Identify all vertical asymptotes of f or g. Show the work that leads to your conclusion. f ( ) (c) Determine lim. Show the work that leads to your conclusion. 0 g ( )

18 Chapter cos 9. Define f( ) = and (a) Show g( ) f ( ) 0. 1 cos 1 1 cos 1 1 cos 0 1 cos g ( ) = on the interval (0, ).