LIMITS AND CONTINUITY

LIMITS AND CONTINUITY 1 The concept of it Eample 11 Let f() = 2 4 Eamine the behavior of f() as approaches 2 2 Solution Let us compute some values of f() for close to 2, as in the tables below We see from the first table that f() is getting closer and closer to 4 as approaches 2 from the left side We epress this by saying that the it of f() as approaches 2 from left is 4, and write f() = 4 Similarly, by looking at the second table, we say that the it of f() as approaches 2 from right is 4, and write f() = 4 + We call f() and f() one-sided its Since the two one-sided its of f() are + the same, we can say that the it of f() as approaches 2 is 4, and write f() = 4 Note that since 2 4 = ( 2)( + 2), we can write 2 4 f() = 2 ( 2)( + 2) = 2 = ( + 2) = 4, where we can cancel the factors of ( 2) since in the it as 2, is close to 2, but 2, so that 2 0 Below, find the graph of f(), from which it is also clear that f() = 4 1

2 LIMITS AND CONTINUITY Eample 12 Let g() = 2 5 Eamine the behavior of g() as approaches 2 2 Solution Based on the graph and tables of approimate function values shown below, observe that as gets closer and closer to 2 from the left, g() increases without bound and as gets closer and closer to 2 from the left, g() decreases without bound We epress this situation by saying that the it of g() as approaches 2 from the left is, and g() as approaches 2 from the right is and write g() =, g() = + Since there is no common value for the one-sided its of g(), we say that the it of g() as approaches 2 does not eists and write g() does not eits Eample 13 Use the graph below to determine f(), f(), f() and f() + 1 Solution It is clear from the graph that f() = 2 and f() = 1 +

LIMITS AND CONTINUITY 3 Since f() f(), f() does not eist It is also clear from the graph that + Since 1 f() = f(), f() = 1 1 1 + 1 Eample 14 (1) Graph 3 + 9 2 9 3 + 9 (2) Evaluate 3 2 9 3 + 9 (3) Evaluate 3 2 9 1 f() = 1 and f() = 1 + Solution (1) Note that f() = 3 + 9 2 9 = 3 3 the graph of y = 1, we obtain for 3 Then, by shifting and scaling (2) Since f() = 3 + 9 2 9 = 3 3 (3) It is seen from the graph that 3 ± 3 + 9 2 9 Eample 15 Evaluate sin Solution From the following tables and the graph 3 + 9 for 3, 3 2 9 = 3 = ± Hence, 3 3 3 = 1 2 3 + 9 2 9 does not eist one can conjecture that sin = 1 From now on, we will use the following fact without giving its proof sin = 1

4 LIMITS AND CONTINUITY Eample 16 Evaluate Solution Note that So, + = 1 while { 1 if > 0 = 1 if > 0 = 1 Since the left it is not equal to the right it, does not eist Eample 17 Sketch the graph of f() = (a) f() (b) f() + (c) f() (d) f() Solution The graph is shown below Eample 18 Sketch the graph of f() = (a) f() (b) f() + (c) f() (d) f() 1 { 2 if < 2 2 if 2 And, (a) (b) and identify each it f() = 2 = 4 f() = 2 = 4 + + (c) f() = 4 (d) f() = 2 = 2 3 1 if < 0 0 if = 0 + 1 2 if > 0 and identify each it

LIMITS AND CONTINUITY 5 (e) 3 f() Solution The graph is shown below And, (a) f() = 3 1 = 1 (b) f() = + + 1 2 = 1 + (c) f() = 1 (d) 1 f() = 1 3 1 = 2 (e) 3 f() = 3 + 1 2 = 0 2 Computation of Limits It is easy to see that for any constant c and any real number a, c = c, and = a The following theorem lists some basic rules for dealing with common it problems Theorem 21 Suppose that f() and g() both eist and let c be any constant Then, (i) [cf()] = c f(), (ii) [f() ± g()] = f() ± g(), [ ] [ ] (iii) [f()g()] = f() g(), and f() f() (iv) g() = provided g() 0 g() By using (iii) of Theorem 21, whenever f() eits, [f()]2 = [f()f()] [ ] [ ] [ ] 2 = f() f() = f() Repeating this argument, we get that [ ] n [f()]2 = f(), for any positive integer n In particular, for any positive integer n and any real number a, Eample 21 Evaluate (1) (3 2 5 + 4) n = a n

6 LIMITS AND CONTINUITY 3 5 + 4 (2) 3 2 2 2 1 (3) 1 Theorem 22 For any polynomial p() and any real number a, p() = p(a) Theorem 23 Suppose that f() = L and n is any positive integer Then, n f() = n f() = n L, where for n even, we assume that L > 0 Eample 22 Evaluate (1) 5 3 2 2 + 2 2 (2) Theorem 24 For any real number a, we have (i) sin = sin a, (ii) cos = cos a, (iii) e = e a, (iv) ln = ln a, for a > 0, (v) sin 1 = sin 1 a, for 1 < a < 1, (vi) cos 1 = cos 1 a, for 1 < a < 1, (vii) tan 1 = tan 1 a, for < a <, (viii) if p is a polynomial and Eample 23 Evaluate sin 1 p(a) Eample 24 Evaluate ( cot ) f() = L, then ( + 1 2 ) Theorem 25 (Sandwich Theorem) Suppose that f() g() h() f(p()) = L for all in some interval (c, d), ecept possibly at the point a (c, d) and that for some number L Then, it follows that f() = h() = L, g() = L, too

[ Eample 25 Evaluate 2 cos LIMITS AND CONTINUITY 7 ( 1 )] Eample 26 Evaluate f(), where f is defined by Eample 27 Evaluate 1 e 2 (1) 1 e 3 1 (2) 2 + 2 3 sin (3) tan tan 2 (4) 5 e 2+1 (5) 2 + (6) 2 csc 2 + (7) ( 1 1 2 2 1 (1 + ) 3 1 (8) sin (9) (10) (11) 5 (12) ( ) f() = ) { 2 + 2 cos + 1 if < 0 e 4 if 0 A function f is continuous at = a when (i) f(a) is defined, (ii) f() eists, and (iii) f() = f(a) 3 Continuity and Its Consequences Otherwise f is said to be discontinuous at = a Eample 31 Let us see some eamples of functions that are discontinuous at = a (1) The function is not defined at = a The graph has a hole at = a

8 LIMITS AND CONTINUITY (2) The function is defined at = a, but f() does not eist The graph has a jump at = a (3) f() eists and f(a) is defined but f() f(a) The graph has a hole at = a (4) f() = and so f() = f(a) never holds The function blows up at = a Eample 32 Determine where f() = 2 + 2 3 is continuous 1 The point = a is called a removable discontinuity of a function f if one can remove the discontinuity by redefining the function at that point Otherwise, it is called a nonremovable or an essential discontinuity of f Clearly, a function has a removable discontinuity at = a if and only if f() eists and is finite

Eample 33 Classify all the discontinuities of (1) f() = 2 + 2 3 1 (2) f() = 1 2 (3) f() = cos 1 LIMITS AND CONTINUITY 9 Theorem 31 All polynomials are continuous everywhere Additionally, sin, cos, tan 1 and e are continuous everywhere, n is continuous for all, when n is odd and for > 0, when n is even We also have ln is continuous for > 0 and sin 1 and cos 1 are continuous for 1 < < 1 Theorem 32 Suppose that f and g are continuous at = a Then all of the following are true: (1) (f ± g) is continuous at = a, (2) (f g) is continuous at = a, and (3) (f/g) is continuous at = a if g(a) 0 Eample 34 Find and classify all the discontinuities of 4 3 2 + 2 2 3 4 Theorem 33 Suppose that g() = L and f is continuous at L Then, ( ) f(g()) = f g() = f(l) Corollary 34 Suppose that g is continuous at a and f is continuous at g(a) composition f g is continuous at a Eample 35 Determine where h() = cos( 2 5 + 2) is continuous Then the If f is continuous at every point on an open interval (a, b), we say that f is continuous on (a, b) We say that f is continuous on the closed interval [a, b], if f is continuous on the open interval (a, b) and b f() = f(a) and f() = f(b) + Finally, if f is continuous on all of (, ), we simply say that f is continuous Eample 36 Determine the interval(s) where f is continuous, for (1) f() = 4 2, (2) f() = ln( 3) Eample 37 For what value of a is continuous at every? f() = { 2 1, < 3 2a, 3

10 LIMITS AND CONTINUITY Eample 38 Let If f is continuous at = 1, find a and b 2 sgn( 1), > 1, f() = a, = 1, + b, < 1 Theorem 35 (Intermediate Value Theorem) Suppose that f is continuous on the closed interval [a, b] and W is any number between f(a) and f(b) Then, there is a number c [a, b] for which f(c) = W Eample 39 Two illustrations of the intermediate value theorem: Corollary 36 Suppose that f is continuous on [a, b] and f(a) and f(b) have opposite signs Then, there is at least one number c (a, b) for which f(c) = 0 4 Limits involving infinity; asymptotes If the values of f grow without bound, as approaches a, we say that f() = Similarly, if the values of f become arbitrarily large and negative as approaches a, we say that f() = A line = a is a vertical asymptote of the graph of a function y = f() if either Eample 41 Evaluate 1 (1) (2) 3 1 ( + 3) 2 ( 2) 2 (3) 2 4 2 (4) 2 4 (5) + 3 2 4 f() = ± or f() = ± +

3 (6) 2 4 3 (7) (8) 2 4 2 ( 2) 3 (9) 5 1 ( 5) 3 + 1 (10) 2 ( 3)( + 2) (11) tan π 2 LIMITS AND CONTINUITY 11 Intuitively, f() = L (or, f() = L if moves increasingly far from the origin in the positive direction (or, in the negative direction), f() gets arbitrarily close to L 1 Eample 42 Clearly, = 0 and 1 = 0 A line y = b is a horizontal asymptote of the graph of a function y = f() if either Eample 43 Evaluate 5 + 1 f() = b or f() = b Theorem 41 For any rational number t > 0, 1 ± = 0, t where for the case where, we assume that t = p/q where q is odd Theorem 42 For any polynomial of degree n > 0, p n () = a n n + a n 1 n 1 + + a 0, we have { if p an > 0 n() = if a n < 0 Eample 44 Evaluate 2 3 + 7 (1) 3 2 + + 7 3 + 7 (2) 2 2 (3) sin 1 (4) 2 + 2 + 3 sin (5) (6) sin Eample 45 Find the horizontal asymptote(s) of f() = 2 + sin + cos

12 LIMITS AND CONTINUITY Let f() = P () If (the degree of P ) = (the degree of Q)+1, then the graph of f has a Q() oblique (slant) asymptote We find an equation for the asymptote by dividing numerator by denominator to epress f as a linear function plus a remainder that goes to 0 as ± Eample 46 Find the asymptotes of the graph of f, if (1) f() = 2 3 2 4 (2) f() = 2 + 1 Eample 47 Evaluate (1) e 1 (2) e 1 + (3) tan 1 (4) tan 1 (5) ln + (6) ln (7) sin (8) (9) (e 1 2 ) 2 + 1 2 + 1