Advanced Level Pure Mathematics

Continuity Advanced Level Pure Mathematics

Advanced Level Pure Mathematics Calculus I 3

Chapter 1 Continuity

3.1 Introduction 2

3.2 Limit of a Function 2

3.3 Properties of Limit of a Function 9

3.4 Two Important Limits 10

3.5 Left and Right Hand Limits 12

3.6 Continuous Functions 13

3.7 Properties of Continuous Functions 16

Prepared by K. F. Ngai Page 1 Continuity Advanced Level Pure Mathematics

3.1 Introduction

1 f (x)  x2 f (x)  Example is continuous on R. Example x is discontinuous at x = 0. y y 1 y  x 2 y  x

0 x 0 x

3.2 Limit of a Function

A. Limit of a Function at Infinity

lim f (x)  ℓ Defintion Let f(x) be a function defined on R. x means that for any  > 0 , there exists X > 0 such that when x > X , f (x)  ℓ   . y y

l+ l+ l l l

0 X x 0 X x

lim f (x)  ℓ N.B.(1) x means that the difference between f(x) and A can be made arbitrarily small when x is sufficiently large. lim f (x)  ℓ (2) x means f(x) → A as x → ∞. (3) Infinity, ∞, is a symbol but not a real value.

There are three cases for the limit of a function when x → ∞, 1. f(x) may have a 2. f(x) may approach to 3. f(x) may oscillate or finite limit value. infinity; infinitely and limit value does not exist.

0 x 0 x

Prepared by K. F. Ngai Page 2 Continuity Advanced Level Pure Mathematics Theorem UNIQUENESS of Limit Value lim f (x)  a lim f (x)  b a  b If x and x , then .

Theorem Rules of Operations on Limits lim f (x) lim g(x) If x and x exist , then lim[ f (x)  g(x)]  lim f (x)  lim g(x) (a) x x x lim f (x)g(x)  lim f (x)  lim g(x) (b) x x x f (x) lim f (x) lim  x lim g(x)  0 x g(x) lim g(x) x (c) x if . lim[kf (x)]  k lim f (x) (d) For any constant k, x x . (e) For any positive integer n, lim[ f (x)]n  [lim f (x)]n (i) x x

lim n f (x)  n lim f (x) (ii) x x

1 lim  0 N.B. x x

2 3x  2x  7 4 2 lim lim( x 1  x ) Example Evaluate (a) x 5x2  x 1 (b) x

Theorem Let f (x) and g(x) be two functions defined on R. Suppose X is a positive real number. f (x) lim g(x)  0 lim f (x)g(x)  0 (a) If is bounded for x > X and x , then x f (x) lim g(x)   lim[ f (x)  g(x)]   (b) If is bounded for x > X and x , then x ; lim f (x)  0 f (x) lim g(x)   (c) If x and is non-zero for x > X , and x , then lim f (x)g(x)   x .

sin x x x lim lim e cos x lim(e  sin x) Example Evaluate (a) x x (b) x (c) x xex x  cos x lim lim (d) x 1 x (e) x x 1

Theorem SANDWICH THEOREM FOR FUNCTIONS Let f(x) , g(x) , h(x) be three functions defined on R. lim f (x)  lim h(x)  a If x x and there exists a positive real number X such that when x > X , f (x)  g(x)  h(x) lim g(x)  a , then x .

y = h(x)

y = f(x) 0 x

(x 1)(x  3)  (x 1)(x  3)  x  2 Example (a) Show that for x > 0, x  2 . lim[ (x 1)(x  3)  x] (b) Hence find x .

1 lim(1 )n  e N.B. n n for any positive integer n.

1 lim(1 )x  e Example Evaluate x x by sandwich rule.

1 1 lim(1 ) x  e  lim(1 y) y  e Theorem x x y0 x2 1  x 2 1 1 lim(1 x) x lim  1x x0  2  lim x Example Evaluate (a) (b) x x 1 (c) x1

1 1 x lim (1 )  e lim(1 x) x  e N.B. 1. x x 2. x0 . 1 2 x lim(1 3x) x lim(1 ) Exercise (a) x0 (b) x x

cot x x 1 lim(1 tan x) lim( ) x (c) x (d) x x 1 Prepared by K. F. Ngai Page 5 Continuity Advanced Level Pure Mathematics

B. Limit of a Function at a Point

lim f (x)  ℓ Definition Let f(x) be a function defined on R. xa means that for any  > 0 , there exists  > 0 such that when 0  x  a   , f (x)  ℓ   . y y = f(x)

f(x) l

0 a x x

lim f (x)  ℓ N.B. (1) xa means that the difference between f(x) and A can be made arbitrarily small when x is sufficiently close to a. lim f (x)  f (a) (2) If f(x) is a polynomial , then xa . lim f (x)  f (a) (3) In general , xa . lim f (x) (4) f(x) may not defined at x = a even through xa exists.

lim(x 2  2x  5)  32  2(3)  5  8 limcos x  cos  1 Example x3 , x .

3 if x  5 f (x)   lim f (x)  3 Example Let 1 if x = 5 , then x5 but f(5) = 1 , lim f (x)  f (5) so x5 .

2x 1, x  2 f (x)   . Example Consider the function  4, x  2 f (x) x  2 lim f (x)  lim(2x 1)  5  f (2). is discontinuous at since x2 x2

x 2  4 (x  2)(x  2) f (x)  lim f (x)  lim  lim(x  2)  4 Example Let x  2 , then x2 x2 x  2 x2 .

Prepared by K. F. Ngai Page 6 Continuity Advanced Level Pure Mathematics But f(2) is not defined on R. lim f (x)  l N.B. xa is equivalent to : lim [ f (x)  ℓ]  0 (i) xa ; lim f (x)  ℓ  0 (ii) xa ; lim f (x)  ℓ  lim f (x); (iii) xa xa lim f (a  h)  ℓ (iv) h0 .

Theorem Rules of Operations on Limits Let f (x) and g(x) be two funcitons defined on an interval containing a , possibly except a . lim f (x)  h lim g(x)  k If xa and xa , then lim f (x)  g(x)  h  k (a) xa lim f (x)g(x)  hk (b) xa f (x) h lim  (if k  0) (c) xa g(x) k ,

x3 1 (1 x)n 1 lim lim Example Evaluate (a) x1 x 1 (b) x0 x 2  x lim lim ( x2  x  x) (c) x2 3  x2  5 (d) x

1 x 1 x lim lim Example Evaluate (a) x3 2  2x 2 (b) x1 2  2x 2 x  2 1 x lim lim (c) x2 x 2  2  2 (d) x1 1 3 x

sin x tan x 2 x (  ,0)  (0,  ) 2   Example (a) Prove that for 2 2 , 1 cos x cos x . sin x tan x lim  2 (b) Hence, deduce that x0 1 cos x .

3.3 Properties of Limit of a Functon

Theorem Uniqueness of Limit Value lim f (x)  h lim f (x)  k h  k If xa and xa , then .

Theorem Heine Theorem

lim f (x)  l  lim f (xn )  l where xn  a as n   xa n .

Prepared by K. F. Ngai Page 8 Continuity Advanced Level Pure Mathematics 1 limsin Example Prove that x0 x does not exist.

x3  ax 2  x  4 lim a Example If x1 x 1 exists, find the value of and the limit value.

Assignment Ex.3A (1-6)

3.4 Two Important Limits

sin x lim  1 Theorem x0 x

sin x cos x x tan x lim  lim  lim  lim  N.B . (1) x x x x (2) x0 sin x x0 x cos x 1 lim  limsin  (3) x0 x (4) x0 x

Example Find the limits of the following functions: tan x 1 cos x lim lim (a) x0 x (b) x0 x 2  sin 4x lim(x  )sec x lim x x0 (c) 2 2 (d) sin 5x

Example Find the limits of the following functions: sin(x 1) cosax  cosbx 3sinx  sin 3x lim lim lim (a) x1 x 2  3x  2 (b) x0 x 2 (c) x0 x3

  0    Example Let be a real number such that 2 .    sin  2sin cos By using the formula 2k 2k1 2k1 , show that θ θ θ θ sinθ  2n1 cos cos ...cos sin 2 22 2n1 2n-1 . x x x lim(cos cos ...cos ) Hence, find the limit value n 2 22 2n

Assignment Ex. 3B (1-2)

3.5 Left and Right Hand Limits

lim f (x)  l iff lim f (x)  lim f (x)  l Theorem xa xa xa .

Example Show that each of the following limits does not exist.

x 1 lim x  2 lim lime x (a) x2 (b) x0 x (c) x0

1 1 lim(a x  b x ) x a  b  0 Example By Sandwich rule, show that x0 does not exist for . b a  b  0 1 a  b  0 x 0 Solution If then a If and then b 1 1 b 1 a 1 x  0 ( ) x 1x 1 ( ) x 1  ( ) x 1 If then a a b 1 1 b 1 1 1 a 1  a  (a x  b x ) x  a{1 ( ) x }x  a  2 x b  (a x  b x ) x  b{( ) x 1}x  b  2 x a b lim 2 x 1 lim b  2 x  b As x0 , by sandwich rule, As x0 , by sandwich rule, 1 1 1 1 lim(a x  b x ) x  a lim(a x  b x ) x  b we have x0 we have x0

1 1 a  b, lim(a x  b x ) x Since x0 does not exist. ( Why ? )

Assignment Ex. 3C (1-3)

3.6 Continuous Functions

lim f (x)  f (a) N.B. In general, xa .

Definition Let f(x) be a function defined on R. f(x) is said to be continuous at x = a if and only if lim f (x)  f (a) xa .

N.B. This is equivalent to the definition : A function f(x) is continuous at x = a , if and only if (a) f(x) is well-defined at x = a , i.e. f(a) exists and f(a) is a finite value, lim f (x) lim f (x)  f (a) (b) xa exists and xa . lim f (a  h)  f (a) Remark Sometimes, the second condition may be written as h0 Prepared by K. F. Ngai Page 12 Continuity Advanced Level Pure Mathematics  1 xsin , x  0 f (x)   x Example Show that the function 0 , x = 0 is continuous at x = 0.

Definition A function is discontinuous at x = a iff it is not continuous at that point a. y There are four kinds of discontinuity: y = f(x) f(a)  (1) Removable discontinuity: o lim f (x) xa exists but not equal to f(a). x 0 a 2x 1, x  2 f (x)   Example Show that 4 , x = 2 is discontinuous at x = 2.

lim f (x)  lim(2x 1)  5  f (2) Solution Since f(2) = 4 and x2 x2 , so f(x) is discontinuous at x = 2.

1 cos x  , x  0 f (x)   x 2 Example Let  a , x = 0 . lim f (x) (a) Find x0 . (b) Find a if f(x) is continuous at x = 0.

y (2) Jump discontinuity :  y = f(x) lim f (x)  lim f (x) xa xa o

Example Show that the function f(x) = [x] is discontinuous at 2.

Example Find the points of discontinuity of the function g(x) = x  [x].

x3  2x 1, x  0 f (x)   Example Let sin x, x  0 . Show that f(x) is discontinuous at x = 0.

(3) Infinite discontinuity: y

lim f (x)   or lim f (x)   xa xa y = f(x) i.e. limit does not exist. 0 a x

1 f (x)  Example Show that the function x is discontinuous at x = 0.

1 (4) At least one of the one-side limit does not exist. f (x)  e x lim f (x) or lim f (x) xa xa  do not exist.

1 1 lim e x  0 and lim e x   x Example Since x0 x0 ,

1 so the function f (x)  e x is discontinuous at x = 0.

Definition (1) A function f(x) is called a continuous function in an open interval (a, b) if it is continuous at every point in (a, b).

(2) A function f(x) is called a continuous function in an closed interval [a, b] if it is continuous at lim f (x)  f (a) lim f (x)  f (b) every point in (a, b) and xa  and xb . (1) (2) y y y

o  a b x a b x a b x f(x) is continuous f(x) is continuous f(x) is continuous on (a, b). on (a, b). on [a, b].

f (x) lim f (x)  f (a)  lim f (a  h)  f (a) N.B. (1) is continuous at x = a  xa h0 . f (x) lim f (x  h)  f (x) (2) is continuous at every x on R  h0 for all x  R. Example Let f (x) be a real-value function such that (1) f (x  y)  f (x) f (y) for all real numbers x and y , (2) f (x) is continuous at x  0 and f (0)  1. Show that f (x) is continuous at every x  R.

3.7 Properties of Continuous Functions

Prepared by K. F. Ngai Page 15 Continuity Advanced Level Pure Mathematics A. Continuity of Elementary Functions Theorem Rules of Operations on Continuous Functions If f(x) and g(x) are two functions continuous at x = a , then so are f (x)  g(x) , f (x)g(x) f (x) and g(x) provided g(a)  0.

1 cos x h(x)  f (x)  1 cos x g(x)  x 2 Example Let x 2 . Since and are two functions continuous everywhere, but g(0) = 0, so h is continuous everywhere except x = 0.

1 cos x h(x)  f (x)  1 cos x g(x)  x 2 1 ( 0) Example Let x 2 1 . Since and are two functions continuous everywhere, so h is continuous everywhere.

Theorem Let g(x) be continuous at x = a and f(x) be continuous at x = g(a), then f。g is continuous at x = a.

x sin x Example sin e , e are continuous functions.

2 Example Show that f (x)  cos x is continuous at every x  R.

x | x |  x if x  0 f (x)  g(x)   2 Example Let f and g be two functions defined as 2 for all x and x if x  0 . For what values of x is f [g(x)] continuous?

lim f (x)  A sin x, cos x, 1 x Theorem Let x and g(x) be an elementary function (such as sin x, e , ln x, ... etc.) for which g(A) is defined, then

lim g[ f (x)]  g[lim f (x)]  g(A) x x

limcos(tanx)  cos(lim tanx)  Example x1 x1

Prepared by K. F. Ngai Page 16 Continuity Advanced Level Pure Mathematics ln(1 x) lim Example Evaluate x x .

sinx lim tan ( ) Example Evaluate x0 x .

B. Properties of Continuous Functions

(P1) If f(x) is continuous on [a, b], then f(x) is bounded on [a, b].

(P2) But f(x) is continuous on (a, b) cannot imples that f(x) is bounded on (a, b).

(1) y c < f(x) < d (2) y f(x) is not bounded d  on (a, b).

c  0 a b x

0 a b x

Example f(x) = cot x is continuous on (0, ) , but it is not bounded on (0, ).

(P3) If f(x) is continuous on [a, b], then it will attain an absolute maximum and absolute minimum on [a, b].

Prepared by K. F. Ngai Page 17 Continuity Advanced Level Pure Mathematics i.e. If f(x) is continuous on [a, b], there exist x1 [a,b] such that f(x1 )  f(x) and x2 [a,b] such

that f(x2 )  f(x) for all x[a,b] . x1 is called the absolute maximum of the function and x2 is called the absolute minimum of the function. no. abs. max. on [a, b] y o abs. max. on [c, d] | o  

0 a b c d    abs. min. on [c, d]  abs. min. on [a, b]

(P4) If f(x) is continuous on [a, b] and f(a)f(b) < 0 , then there exists c  [a, b] such that f(c) = 0.

 

0 a b x 0 a b x     Example Let f (x)  cos x  0.6 which is continuous on [0, 2 ] and f(0) = 0.4 and f ( 2 )  0.6 . Hence,  there exists a real number c  [0, 2 ] such that f(c) = 0.

Example Prove that the equation x e x  2 has at least one real root in [0, 1].

3 Example Let f (x)  x  2x .

Prepared by K. F. Ngai Page 18 Continuity Advanced Level Pure Mathematics (a) Show that f(x) is a strictly increasing function. (b) Hence, show that the equation f (x)  0 has a unique real root.

(P5) Intermediate Value Theorem If f(x) is continuous on [a, b], then for any real number m lying between f(a) and f(b) , there corresponds a number c  [a, b] such that f(c) = m. y

f(b)  m

f(a)  0 a c b x

1 (P6) Let f(a) = c and f(b) = d, if f is continuous and strictly increasing on [a, b], then f is also continuous and strictly increasing on [c, d] ( or [d, c] ).

2 1 Example f (x)  x is strictly increasing and continuous on [0, 5], so f (x)  x is also strictly increasing and continuous on [0, 25].

1 1 Example f (x)  cos x is strictly decreasing and continuous on [0,π], so f (x)  cos x is also strictly decreasing and continuous on [1, 1].

Prepared by K. F. Ngai Page 19 Continuity Advanced Level Pure Mathematics Example (a) Suppose that the function satisfies f(x+y) = f(x) + f(y) for all real x and y and f(x) is continuous at x = 0. Show that f(x) is continuous at all x.  1 sin x  1 sin x  if x  0 f (x)   x (b) A function  0 if x = 0 . Is f(x) continuous at x = 0? If not , how can we redefine the value of f(x) at x = 0 so that it is continuous at x = 0?

Exercise Example Let f(x) be a continuous function defined for x > 0 and for any x , y > 0, f(xy) = f(x) + f(y). (a) Find f(1).

r (b) Let a be a positive real number. Prove that f (a )  r  f (a) for any non-negative rational number.

(c) It is known that for all real numbers x, there exists a sequence {xn} of rational numbers such

lim xn  x that x . x (i) Show that f (a )  x  f (a) for all x > 0, where a is a positive real constant. (ii) Hence show that f (x)  c ln x for all x > 0, where c is a constant.

Example Let f be a real-valued continuous function defined on the set R such that f(x+y) = f(x) + f(y) for all x , y  R. (a) Show that (i) f(0) = 0 (ii) f(x)= f(x) for any x  R. (b) Prove that f(nx) = nf(x) for all integers n. Hence show that f(r) = rf(1) for any rational number r.

(c) It is known that for all x  R, there exists a sequence {an }of rational numbers such that

lim an  x n . Using (b), prove that there exists a constant k such that f(x) = kx for all x  R.

Example Let R denote the set of all real numbers and f : R  R a continuous function not identically zero such that f(x+y) = f(x)f(y) for all x , y  R. (a) Show that (i) f(x)  0 for any x  R. (ii) f(x) > 0 for any x  R. (iii) f(0) = 1 1 (iv) f (x)   f (x) r (b) Prove that for any rational number r, f (rx)   f (x) . x Hence prove that there exists a constant a such that f (x)  a for all x  R.

Assignment Ex.3D (1-9). Revision Exercise (1-6).

Prepared by K. F. Ngai Page 22