MA4001 Engineering Mathematics 1 Lecture 10 Limits and Dr. Sarah Mitchell Autumn 2014
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.
Example f(x) = 1 x 1 lim x 0+ x = lim 1 x 0 x = 1 Therefore lim does not exist. x 0 x
Example f(x) = x 2 + 1 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.
Graph of x 2 + 1 x 2 1
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.
Remark If f(x) is not defined at c, then f(x) is neither continuous nor discontinuous at c.
Example f(x) = x 2 + 1 lim f(x) = x 1 12 + 1 = 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.
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.
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) = 1 2 + 1 = 2 f(1) = 1 x 1 x 1
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.
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.
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
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
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.
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].
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, ).
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
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.
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}.
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.
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.
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].
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 ) }.