MA4001 Engineering Mathematics 1 Lecture 10 Limits and Continuity

Transcription

1 MA4001 Engineering Mathematics 1 Lecture 10 Limits and Dr. Sarah Mitchell Autumn 2014

2 Infinite limits If f(x) grows arbitrarily large as x a we say that f(x) has an infinite limit. Example: f(x) = 1 x 2. lim x 0 f(x) = The line x = 0 is a vertical asymptote.

3 Example f(x) = 1 x 1 lim x 0+ x = lim 1 x 0 x = 1 Therefore lim does not exist. x 0 x

4 Example f(x) = x x 2 1 Clearly the interesting x values are ±1 and ±. Note that f( x) = f(x), so the function is even. Thus we only need to check the behaviour at 1 and. lim f(x) = + lim x 1+ f(x) = (since x 2 1 < 0 for x < 1) x 1 1+1/x 2 lim f(x) = lim x x 1 1/x 2 = 1 Thus x = ±1 are vertical asymptotes and y = 1 is a horizontal asymptote. Also f(0) = 1.

5 Graph of x x 2 1

6 at a point Definition f(x) is continuous at c, if If c is an interior point of the domain of f lim f(x) = f(c) i.e., the graph of f(x) is continuous at c. x c lim f(x) f(c) or x c lim x c f(x) does not exist then we say that f(x) is discontinuous at c, i.e., the graph is discontinuous at c.

7 Remark If f(x) is not defined at c, then f(x) is neither continuous nor discontinuous at c.

8 Example f(x) = x lim f(x) = x = 2 = f(1) Thus f(x) is continuous at 1. We could repeat this for any value of x, so in fact f(x) is continuous in R.

9 The Heaviside function { 1, x 0 H(x) = is continuous in R\{0}. 0, x < 0 H(x) is discontinuous at x = 0 since lim x 0 H(x) doesn t exist.

10 Example: Removable Singularity f(x) = { x 2 + 1, if x 1 1, if x = 1 f(x) is continuous on R\{1}. f(x) is discontinuous at x = 1 since lim f(x) = lim (x 2 + 1) = = 2 f(1) = 1 x 1 x 1

11 Removable Singularity Definition f(x) has a removable singularity at x = c if f(x) can be made continuous at c by redefining f(c) to be f(c) = lim x c f(x) In the previous example, f(x) had a removable singularity at x = 1. By redefining f(1) = 2 the function is made continuous at x = 1.

12 Example sgn(x) sgn(x) is continuous on its domain R\{0}. At x = 0, sgn(x) is not defined. Thus it is neither continuous nor discontinuous at 0.

13 Left and right continuity Definition f(x) is right continuous at x = c if lim f(x) = f(c). x c+ f(x) is left continuous at x = c if lim f(x) = f(c). x c

14 Example: Heaviside function { 1, x 0 H(x) = is right continuous at x = 0 since 0, x < 0 lim H(x) = 1 = H(0) x 0+ but not left continuous at x = 0 since lim H(x) = 0 1 = H(0) x 0

15 at endpoints of domains Definition f(x) is continuous at a left endpoint of its domain, if it is right continuous at this point. f(x) is continuous at a right endpoint of its domain, if it is left continuous at this point.

16 Example f(x) = 1 x 2 The domain is 1 x 2 0, i.e., x 2 1, i.e., [ 1, 1] f(x) is continuous on ( 1, 1). f(x) is right continuous at x = 1 and left continuous at x = 1. Thus f(x) is continuous on [ 1, 1].

17 Example f(x) = x 2 The domain is x 2 0, i.e., x 2, i.e., [2, ) f(x) is continuous on (2, ). f(x) is right continuous at x = 2 Thus f(x) is continuous on its whole domain [2, ).

18 on R Many functions are continuous on R. These are referred to as continuous functions. Examples: all polynomials; all rational functions with non-zero denominator; sin x, cos x, e x

19 Combinations of continuous functions If f(x) and g(x) are continuous at c then the following are continuous at c: f(x)+g(x) f(x) g(x) f(x)g(x) f(x) g(x) provided g(c) 0.

20 Example sin x, x, e x are all continuous on R. Then 3 sin x e x + 8 x is continuous on its domain R\{0}.

21 of composite functions (f g)(x) = f(g(x)) If g(x) is continuous at c and f(x) is continuous at g(c) then (f g)(x) is continuous at c. Example. f(x) = x, g(x) = x 2 2x + 5. (f g)(x) = x 2 2x + 5 is continuous on R. Note that the domain of f g is R.

22 Continuous functions on [a, b] Recall that f(x) is continuous on the closed interval [a, b] if f(x) is continuous at each x (a, b); f(x) is right continuous at x = a; f(x) is left continuous at x = b.

23 The max-min theorem Theorem If f(x) is continuous on [a, b], there exist x 1, x 2 [a, b], such that f(x 1 ) f(x) f(x 2 ), x [a, b] We say that f(x) has the absolute maximum M = f(x 2 ) at x = x 2 and the absolute minimum m = f(x 1 ) at x = x 1 on [a, b].

24 Corollary If f(x) is continuous on [a, b], then it is bounded on [a, b] i.e., there exists K 0 such that f(x) K for all x [a, b]. Proof. Choose K = max{ f(x 1 ), f(x 2 ) }.