2 Continuity, Differentiability and Taylor's theorem 2.1 Limits of real valued functions Let f(x) be defined on (a; b) except possibly at x0. Definition 2.1.1. We say that lim f(x) = L if, for every real number > 0; there exists a real x!x0 number δ > 0 such that 0 < jx − x0j < δ =) jf(x) − Lj < . (2.1) Equivalently, Remark 2.1. The above definition is equivalent to: for any sequence fxng with xn ! x0, we have f(xn) ! L as n ! 1. Proof. Suppose limx!x0 f(x) exists. Take > 0 and let fxng be a sequence converging to x0. Then there exists N such that jxn − x0j < δ for n ≥ N. Then by the definition jf(xn) − Lj < . i.e., f(xn) ! L. For the other side, assume that xn ! c =) f(xn) ! L. Suppose the limit does not exist. i.e., 9 0 > 0 such that for any δ > 0; there is x 2 jx − x0j < δ, for which jf(x) − Lj ≥ 0. Then take 1 1 δ = n and pick xn in jxn − x0j < n , then xn ! x0 but jf(xn) − Lj ≥ 0. Not possible. Theorem 2.1.2. If limit exists, then it is unique. Proof. Proof is easy. 3x 1 Examples 2.1.3. 1. lim( − 1) = . Let > 0. We have to find δ > 0 such that (2.1) x!1 2 2 holds with L = 1=2. Working backwards, 3 2 jx − 1j < when ever jx − 1j < δ := . 2 3 ( x2 x 6= 2 2. Prove that lim f(x) = 4, where f(x) = x!2 1 x = 2 1 Problem: Show that lim sin( ) does not exist. x!0 x 1 1 Consider the sequences fxng = f g; fyng = f π g. Then it is easy to see that xn; yn ! 0 nπ 2nπ + 2 1 1 and sin ! 0; sin ! 1. In fact, for every c 2 [−1; 1]; we can find a sequence zn such xn yn 1 that zn ! 0 and sin( ) ! c as n ! 1: zn By now we are familiar with limits and one can expect the following: 1 Theorem 2.1.4. Suppose lim f(x) = L and lim g(x) = M, then x!c x!c 1. lim(f(x) ± g(x)) = L ± M. x!c 2. f(x) ≤ g(x) for all x in an open interval containing c. Then L ≤ M. f L 3. lim(fg)(x) = LM and when M 6= 0, lim (x) = : x!c x!c g M 4. (Sandwich): Suppose that h(x) satisfies f(x) ≤ h(x) ≤ g(x) in an interval containing c, and L = M. Then lim h(x) = L. x!c Proof. We give the proof of (iii). Proof of other assertions are easy to prove. Let > 0. From the definition of limit, we have δ1; δ2; δ3 > 0 such that 1 jx − cj < δ =) jf(x) − Lj < =) jf(x)j < N for some N > 0; 1 2 jx − cj < δ =) jf(x) − Lj < ; and 2 2M jx − cj < δ =) jg(x) − Mj < : 3 2N Hence for jx − cj < δ = minfδ1; δ2; δ3g, we have jf(x)g(x) − LMj ≤ jf(x)g(x) − f(x)Mj + jf(x)M − LMj ≤ jf(x)jjg(x) − Mj + Mjf(x) − Lj < . To prove the second part, note that there exists an interval (c − δ; c + δ) around c such that g(x) 6= 0 in (c − δ; c + δ). Examples 2.1.5. (i) lim xm = 0 (m > 0): (ii) lim x sin x = 0: x!0 x!0 Remark 2.2. Suppose f(x) is bounded in an interval containing c and lim g(x) = 0. Then x!c lim f(x)g(x) = 0: x!c 1 Examples 2.1.6. (i) lim jxj sin = 0. (ii) limx!0 jxj ln jxj = 0: x!0 x One sided limits: Let f(x) is defined on (c; b): The right hand limit of f(x) at c is L, if given > 0, there exists δ > 0, such that 0 < x − c < δ =) jf(x) − Lj < . Notation: lim f(x) = L. Similarly, one can define the left hand limit of f(x) at b and is denoted x!c+ by lim f(x) = L: x!b− Both theorems above holds for right and left limits. Proof is easy. 2 sin θ Problem: Show that limθ!0 θ = 1. Solution: Consider the unit circle centered at O(0; 0) and passing through A(1; 0) and B(0; 1). Let Q be the projection of P on x−axis and the point T is such that A is the projection of T on x−axis. Let OT be the ray with \AOT = θ; 0 < θ < π=2. Let P be the point of intersection of OT and circle. Then ∆OP Q and ∆OT A are similar triangles and hence, Area of ∆OAP < Area of sector OAP < area of ∆OAT . i.e., 1 1 1 sin θ < θ < tan θ 2 2 2 sin θ sin θ dividing by sin θ, we get 1 > > cos θ. Now lim cos θ = 1 implies that lim = 1. θ θ!0+ θ!0+ θ sin θ Now use the fact that is even function. θ At this stage, it is not difficult to prove the following: Theorem 2.1.7. limx!a f(x) = L exists () limx!a+ f(x) = limx!a− f(x) = L. Limits at infinity and infinite limits Definition 2.1.8. f(x) has limit L as x approaches +1, if for any given > 0, there exists M > 0 such that x > M =) jf(x) − Lj < . Similarly, one can define limit as x approaches −∞. 1 1 Problem: (i) lim = 0, (ii) limx→−∞ = 0: (iii) lim sin x does not exist. x!1 x x x!1 π Solution: (i) and (ii) are easy. For (iii), Choose xn = nπ and yn = 2 +2nπ. Then xn; yn ! 1 and sin xn = 0, sin yn = 1. Hence the limit does not exist. Above two theorems on limits hold in this also. Definition 2.1.9. (Horizontal Asymptote:) A line y = b is a horizontal asymptote of y = f(x) if either lim f(x) = b or lim f(x) = b. x!1 x→−∞ 1 Example 2.1.10. (i) y = 1 is a horizontal asymptote for 1 + x+1 Definition 2.1.11. (Infinite Limit): A function f(x) approaches 1 (f(x) ! 1) as x ! x0 if, for every real B > 0, there exists δ > 0 such that 0 < jx − x0j < δ =) f(x) > B: Similarly, one can define for −∞. Also one can define one sided limit of f(x) approaching 1 or −∞. 3 1 1 1 Examples 2.1.12. (i) lim = 1, (ii) limx!0 2 sin( ) does not exist. x!0 x2 x x For (i) given B > 0, we can choose δ ≤ p1 : For (ii), choose a sequence fx g such that B n 1 sin 1 = 1, say 1 = π + 2nπ and 1 = nπ: Then lim f(x ) = ! 1 and lim f(y ) = 0; xn xn 2 yn n 2 n n!1 xn n!1 though xn; yn ! 0 as n ! 1. Definition 2.1.13. (Vertical Asymptote:) A line x = a is a vertical asymptote of y = f(x) if either lim f(x) = ±∞ or lim f(x) = ±∞: x!a+ x!a− x+3 Example 2.1.14. f(x) = x+2 . x = −2 is a vertical asymptote and y = 1 is a horizontal asymptote. 2.2 Continuous functions Definition 2.2.1. A real valued function f(x) is said to be continuous at x = c if (i) c 2 domain(f) (ii) lim f(x) exists x!c (iii) The limit in (ii) is equal to f(c). In other words, for every sequence xn ! c, we must have f(xn) ! f(c) as n ! 1. That is, for a given > 0, there exists δ > 0 such that jx − cj < δ =) jf(x) − f(c)j < . ( x2 sin( 1 ) x 6= 0 Examples 2.2.2. 1. f(x) = x is continuous at 0. 0 x = 0 p Let > 0. Then jf(x) − f(0)j ≤ jx2j. So it is enough to choose δ = . ( 1 sin 1 x 6= 0 2. g(x) = x x is not continuous at 0. 0 x = 0 Choose 1 = π + 2nπ. Then lim x = 0 and f(x ) = 1 ! 1: xn 2 n n xn The following theorem is an easy consequence of the definition. Theorem 2.2.3. Suppose f and g are continuous at c. Then (i) f ± g is also continuous at c (ii) fg is continuous at c f (iii) is continuous at c if g(c) 6= 0. g Theorem 2.2.4. Composition of continuous functions is also continuous i.e., if f is continuous at c and g is continuous at f(c) then g(f(x)) is continuous at c. Corollary 2.2.5. If f(x) is continuous at c, then jfj is also continuous at c. Theorem 2.2.6. If f; g are continuous at c, then max(f; g) is continuous at c. 4 Proof. Proof follows from the relation 1 1 max(f; g) = (f + g) + jf − gj 2 2 and the above theorems. Types of discontinuities Removable discontinuity: f(x) is defined every where in an interval containing a except at x = a and limit exists at x = a OR f(x) is defined also at x = a and limit is NOT equal to function value at x = a.