Math 31A Discussion Session Week 1 Notes January 5 and 7, 2016

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1 Math 31A Discussion Session Week 1 Notes January 5 and 7, 2016 This week we re discussing two important topics: its and continuity. We won t give a completely rigorous definition of either, but we ll develop some tools for computing its and determining whether or not functions are continuous. Limit Laws We begin by recalling some basic its. Proposition 1. For real numbers a and b, b = b, a = a, a 1 0 =, = The following it laws, which we state informally, will be a vital part of our toolbo when computing its. Proposition 2. In each of the following rules, we assume that the relevant its eist. 1. The it of a sum is the sum of the its. 2. The it of a difference is the difference of the its. 3. The it of a product is the product of the its. 4. The it of a quotient is the quotient of the its, as long as the it of the denominator is not zero. Eample. From these basic its and simple rules, which all seem obvious enough, we can compute its of any polynomials. Let p() := a n n + a n 1 n a 1 + a 0 be a polynomial, and choose a R. Then To see this, we apply our rules to find that p() = p(a). a p() = (a ( n n ) + an 1 n 1) + + (a 1 ) + (a 0 ) a a a a a = a n n + a n 1 n a 1 + a 0 a a a a ) n ( ) n 1 = a n ( + an a1 + a 0 a a a = a n a n + a n 1 a n a 1 a + a 0 = p(a). 1

2 Net, we have a it law whose statement might sound complicated, but which is easy to use. Proposition 3. If a function f() is non-zero when is near the real number a (but not necessarily nonzero at a), then f()g() a f() = a g(). Note. In general, you won t find yourself worrying about the non-zero near a bit of this rule. It s primarily meant to keep us from concluding that 0 a 0 = 0 g() a 0 = a g() for some function g(), and this is certainly a conclusion we want to avoid. Eample Evaluate 2, if this it eists. 2 (Solution) According to our previous rule (and some factoring), we have = 2 ( 2)( + 3) Evaluate 0, if this it eists = 2 ( + 3) = 5. (Solution) Notice that we can t use rule 4 from Proposition 2 just yet, because we immediately see that the denominator approaches 0 as approaches 0. We must first do some cancellation. We have = 2 ( ) ( ) = = 0( ) 0 ( ) = 1 7. sin 2 (θ) 3. Evaluate θ 0, if this it eists. 1 cos(θ) (Solution) We can use the identity sin 2 (θ) + cos 2 (θ) = 1 to find θ 0 sin 2 (θ) 1 cos(θ) = 1 cos 2 (θ) θ 0 1 cos(θ) = (1 cos(θ))(1 + cos(θ)) θ 0 1 cos(θ) = (1 + cos(θ)) = 2. θ 0 2

3 4. Evaluate 0, if this it eists. (Solution) First, let s rewrite /. We have = { 1, < 0 1, > 0, and / is undefined when = 0. So we see that 0 = 1 = 1 and 0 Since the left and right its do not agree, 0 5. Evaluate 1 1 1, if this it eists. 0 + = 1 = does not eist. (Solution) Since both the numerator and denominator approach 0 as approaches 1, we must find a way to rewrite the fraction. We do this by rationalizing the denominator: 1 = = ( 1)( + 1) Then, using Proposition 3, Continuity = 1 ( 1)( + 1) 1 = 1 ( + 1) = 2. With its, we are able to compute the value that an epression or function should take, if it s nice. Of course, there are many cases where an epression or function doesn t take the value that we think it should, so we distinguish the better-behaved functions by giving them a special name. Definition. We say that a function f is continuous at a point = a if: 1. f(a) is defined; 2. a f() eists; 3. a f() = f(a). We say that f is continuous on the interval (c, d) if f is continuous at each point a (c, d). Eample. 3

4 1. The function f() := is continuous at = 0, but the function {, 0, f() := 42, = 0 is not. (Why?) 2. The function g() := { sin(1/), 0, k, = 0 is continuous on (, 0) and (0, + ), but is not continuous at = 0, no matter the value of k. To see this, let n = 2 (2n + 1)π and y n = 1 nπ. Then g( n ) = sin((2n + 1)π/2) = 1 for all n, and g(y n ) = sin(nπ) = 0 for all n. Since n 0 and y n 0, but g( n ) and g(y n ) approach distinct values, we see that 0 g() does not eist, and thus g cannot be continuous at = 0. As with its, we have some important arithmetic properties for continuous functions: Proposition 4. If f and g are continuous at = a, then 1. f + g is continuous at = a; 2. f g is continuous at = a; 3. f g is continuous at = a; 4. f/g is continuous at = a, provided g(a) 0. Eample. 1. The converses of the above properties are not true. In particular, we can find functions f and g which are discontinuous at = a, but such that f + g, f g, fg, or f/g is continuous. For eample, define f() := { 0, < 0 1, 0 and g() := { 1, < 0 0, 0. Neither of these functions is continuous at = 0, but h() := f()+g() is continuous on all of R, since h 1. We finish with a very useful property of continuous functions: they play nicely with its. This shouldn t be surprising, since we defined continuous functions to be those functions which behave the way our study of its tells us that they should. 4

5 Proposition 5. If f is continuous at = g(a) and a g() eists, then ( ) f(g()) = f g(). a a Eample. The function f() = is continuous on its domain, so = ( 2 + 7) = 16 = 4. 3 Other Eamples 3 Eample. All but the first of these eamples were written by Steven Heilman. 1. Find constants b and c so that the function 4 + b, < 2 f() := c 2, , 2 < is continuous. (Solution) Certainly f is continuous when 2 and 2; we need to make f continuous at these two points as well. First, let s make 2 f() = 2 + f(). We have f() = c2 = 4c and f() = 12 = 12, so in order for 2 f() to eist and equal f(2), we need c = 3. Then But f() = 32 = f() = 2 2 (4 + b) = b 8, so in order for 2 f() to eist and equal f( 2), we must have b = 20. ( 1 2. Evaluate t 0 t 1 ), if this it eists. t 2 + t (Solution) Notice that we cannot split this into a difference of two its, since neither t 0 1/t nor t 0 1/(t 2 + t) eists. Instead, we first combine the fractions: 1 t 1 t 2 + t = t + 1 t(t + 1) 1 t(t + 1) = t t(t + 1) = 1 t + 1. Having made this simplification, we have ( 1 t 0 t 1 ) 1 = t 2 + t t 0 t + 1 = 1. 5

6 3. Evaluate the following it, if it eists (Solution) As with one of our above eamples, this it looks like 0/0, so we ll multiply both the numerator and denominator by the conjugate of the denominator: = = ( ) = This lead us to = = Is there a real number a such that the following it eists? a + a If so, find the value of a and the value of the it. (Solution) Notice that 2 (2 + 2) = 0, so in order for this it to eist, we ll need + 2 to be a factor of the numerator. This will allow us to cancel + 2 from both the numerator and denominator and evaluate the it. To this end, notice that a + a + 3 = ( + 2)(3 + (a 6)) + 15 a. (A little polynomial division will give you this epression.) Thus, if a = 15, we ll have a + a + 3 = ( + 2)(3 + (a 6)), making + 2 a factor of our numerator. Then a + a = 2 ( + 2)( 1) = 2 = 3 3 = 1. 3( + 2)( + 3) ( + 2)( 1) = 2 3( + 3) 1 6