Topic 2 Limits and Continuity c and d) Continuity Handout Notes Assigned Problems: Intro book pg 73, 1-3 and 6-8

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1 c&d. Continuity Handout. Page 1 of 5 Topic Limits and Continuity c and d) Continuity Handout Notes Assigned Problems: Intro book pg 73, 1-3 and 6-8 Recall Limits and Function Values: We have already studied all the concepts necessary to understand continuity To recap, let s look at the following graphs and answer three questions Does f(a) exist? Does lim f ( x ) exist? Does lim f ( x ) = f ( a )? x a x a 1) ) 3) 4) 5) For a function to be continuous, it has to pass all three tests above, if it fails even one it is NOT continuous. Which of the above functions are continuous at a? Informal Definition of Continuity: A function f is continuous at x = a if and only if 1. f(a) is defined. lim f ( x ) exists x a 3. lim f ( x ) = f ( a ) x a If a function f is not continuous at x = a, it is said to be discontinuous at x = a Types of Discontinuities: If the limit of the function exists (and is finite) at x = a, then continuity fails because either f(a) is not defined, or lim f ( x ) f ( a ), the discontinuity is classified as removable. This is because by simply x a redefining f(a) we can make f continuous at a. Which of the above pictures has a removable discontinuity? How would you fix this? If the limit of the function is infinite, then it is called an infinite discontinuity. Which of the above pictures has an infinite discontinuity? If the limit from the right and left of a are not the same, then the function is said to have a jump discontinuity

2 c&d. Continuity Handout. Page of 5 Continuous on an Interval: A function can be continuous from the right and left at a number (basically you only look at a right or left hand limit) A function can be continuous on an interval if it is continuous for all the points in that interval A function is said to be continuous if it is continuous on its entire domain While the same word (i.e. continuous) is used to describe all of these situations, it is important to keep in mind what you are referring to. Here they are from least strict to most strict continuous at a from the left (or right) continuous at a continuous on an interval continuous The following functions are continuous on their domains: polynomials, rational functions, root functions, trig functions, inverse trig functions, exponential functions, log functions. But wait, I thought log functions were not defined for negative values? How could, say log(x) satisfy the above statement if it isn t even defined for x < 0? Formal Definition of Continuity: A function f : D Re is continuous at a point c if and only if ε > 0 δ > 0 s.t. for x D and x c < δ then f( x) f( c) <ε For a function to be continuous, it must be continuous for all points in its domain Proving Continuity at a Point: Example 1: Show f(x) = x + 1 is continuous at x =. Let x < δ for δ > 0 Then f( x) f() = x+ 1 (4+ 1) = x 4 = x < δ So f( x) f() < ε for ε δ < ε i.e. ε > 0 δ > 0 δ < s.t. for x D and x < δ then f( x) f() < ε

3 Example : Show f ( x) = x f : Re Re is continuous at x = Let x < δ for δ > 0 Then f x f x x ( ) () = = 4 Let s assume this is true (for now) so we can find delta ε < x 4 < ε < x < + ε 4 ε 4 4 ε < x < 4+ ε ε < 4 4 ε < x < 4+ ε So take δ = min( 4 ε, 4 +ε ) In fact, further investigation shows that δ = 4+ ε. ( ) c&d. Continuity Handout. Page 3 of 5 ε > 0 ( wlog.... ε < 4) δ > 0 δ = 4 + ε s.t. for x D and x < δ then f( x) f() <ε. Actually, let s plug in some values to see how this works e d = sqrt(4+e)- - d/ f( - d/) - f() < e? + d/ f( + d/) - f() < e? Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Just so you can see there is more than one way to prove Let x < δ for δ > 0 Then since a+ b a + b, we have x+ x + f ( x) = x f : Re Re is cts at x = And since a b a b, we have that x x < δ or δ < x < + δ If we let δ =, then x+ x + < 4+ =6 So, = = + <. So let δ = min(, ε / 6) f( x) f() x x x 6δ This requires more thought but is actually much cleaner

4 Example 3: Show f( x) = x f :[0, ) Re is cts at x = 1 Let x 1 < δ for δ > 0 x + 1 x 1 f( x) f(1) = x 1 = ( x 1) = x + 1 x Now, x 0 x 0 x x + 1 x 1 So, f( x) f(1) < δ 1 ε > 0 δ > 0 δ = ε s.t. for x D and x 1 < δ then f( x) f(1) <ε. Therefore, ( ) Again, values to see e d = e 1 - d/ f(1 - d/) - f(1) < e? 1 + d/ f(1 + d/) - f(1) < e? Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y 0 1 Y Y c&d. Continuity Handout. Page 4 of 5

5 Showing Continuous on a Domain: Example 4: Show f( x) = x f :Re Re is continuous Let x c < δ for δ > 0 and c Re Then since a+ b a + b, we have x + c x + c And since a b a b, we have that x c x c < δ or c δ < x < c + δ If we let δ = c, then x + c x + c < c + c = 3c. So, f ( x) f( c) x c x c x c δ 3c = = + <. So let δ = min( c, ε / c). But this means c = 0 is a special case. If c = 0, then x Therefore, < δ, and ( ) c&d. Continuity Handout. Page 5 of 5 f( x) f(0) = x = x < δ. ε > 0 δ > 0 δ = min( c, ε / c) when c 0, δ = ε when c= 0 s.t. for x, c Re and 0 < x c < δ then f( x) f( c) < ε.