MAT137 Calculus! Lecture 5 Today: 2.5 The Pinching Theorem; 2.5 Trigonometric Limits. 2.6 Two Basic Theorems. 3.1 The Derivative Next: 3.2-3.6 DIfferentiation Rules Deadline to notify us if you have a conflict with Test 1: June 05
Continuity at a point Definition (Continuity at a point) Let f be a function defined at least on an open interval (c p, c + p). We say that f is continuous at c if lim f (x) = f (c). x c
Discontinuity Types Removable Discontinuity c Jump Discontinuity c Infinite Discontinuity c
Continuous Functions are Cool! If f is continuous, 1 To compute the limit we just need to plug-in the value lim f (x) = f (a) x a 2 We can take the limit inside the function. Theorem Let f and g be functions such that THEN lim g(x) = L, and x c f is continuous at L. ( ) lim f (g(x)) = f lim g(x). x c x c
The Pinching Theorem ( a.k.a. Squeeze Theorem) Theorem Let p > 0. Suppose that for all x with 0 < x a < p, f (x) g(x) h(x). IF THEN lim f (x) = L and lim h(x) = L, x a x a lim g(x) = L. x a
The Pinching Theorem ( a.k.a. Squeeze Theorem) Show that ( ) 1 lim x 2 sin = 0. x 0 x
The Pinching Theorem ( a.k.a. Squeeze Theorem) Show that ( ) 1 lim x 2 sin = 0. x 0 x
The Pinching Theorem ( a.k.a. Squeeze Theorem) Show that ( ) 1 lim x 2 sin = 0. x 0 x
The Pinching Theorem ( a.k.a. Squeeze Theorem) Example IF f (x) is bounded and lim x a g(x) = 0, THEN lim x a f (x)g(x) = 0.
The Pinching Theorem ( a.k.a. Squeeze Theorem) Example IF f (x) is bounded and lim x a g(x) = 0, THEN lim x a f (x)g(x) = 0. Exercise IF lim x a f (x) = 0, THEN lim x a f (x) = 0. Hint: First show that f (x) f (x) f (x).
The Pinching Theorem ( a.k.a. Squeeze Theorem) Example Show that the function given by { x is continuous at 0. x Q 0 x / Q
Trigonometric Limits Proposition lim sin(x) = 0 and lim cos(x) = 1 x 0 x 0 Read proof in pages 92-93. I will post some material with the proof in the course website.
Trigonometric Limits Proposition lim sin(x) = 0 and lim cos(x) = 1 x 0 x 0 Read proof in pages 92-93. I will post some material with the proof in the course website. Theorem The sine and cosine functions are everywhere continuous; i.e. for all real numbers c, lim sin(x) = sin(c) and lim cos(x) = cos(c) x c x c
Two Useful Limits Useful Limits lim x 0 sin x x = 1 and lim x 0 1 cos x x = 0. Read proof on page 93. I will post some material with the proof in the course website.
Two Useful Limits Useful Limits lim x 0 sin x x = 1 and lim x 0 1 cos x x = 0. Read proof on page 93. I will post some material with the proof in the course website. In general, lim x 0 sin ax ax = 1 and lim x 0 1 cos ax ax = 0.
Trig. Limits - Example 1 sin(2x) Find lim. x 0 3x Answer: 1 3 2 2 0 3 2 3 Useful Limits lim x 0 sin ax ax = 1 and lim x 0 1 cos ax ax = 0.
Trig. Limits - Example 2 Find lim x 0 sin 2 (x 2 ) 3x 4 cos 5x. Answer: 1 3 2 0 3 1 3 Useful Limits lim x 0 sin ax ax = 1 and lim x 0 1 cos ax ax = 0.
Intermediate Value Theorem (IVT) Theorem (Intermediate Value Theorem) IF f be continuous on [a, b] and K is any number between f (a) and f (b), THEN there is at least one number c between a and b for which f (c) = K.
Intermediate Value Theorem (IVT) Theorem (Intermediate Value Theorem) IF f be continuous on [a, b] and K is any number between f (a) and f (b), THEN there is at least one number c between a and b for which f (c) = K.
IVT - Example 1 Show that there is a solution to the equation x 2 = sin x + 2 cos x on the interval (0, π 2 ). Theorem (Intermediate Value Theorem) IF f be continuous on [a, b] and K is any number between f (a) and f (b), THEN there is at least one number c between a and b for which f (c) = K.
IVT - Example 2 Consider the function f (x) = 4 x We have f ( 1) = 4 < 0 and f (1) = 4 > 0. Can we conclude that f (c) = 0 for some c ( 1, 1)?
IVT - Example 2 Consider the function f (x) = 4 x We have f ( 1) = 4 < 0 and f (1) = 4 > 0. Can we conclude that f (c) = 0 for some c ( 1, 1)? No Theorem (Intermediate Value Theorem) IF f be continuous on [a, b] and K is any number between f (a) and f (b), THEN there is at least one number c between a and b for which f (c) = K.
IVT - Example 3 Example (Fixed point property) Show that if f is continuous on [0, 1] and 0 f (x) 1 for all x [0, 1], then there exists a point p [0, 1] such that f (p) = p.
Extreme Value Theorem (EVT) Theorem (Extreme-Value Theorem) IF f is continuous on a bounded closed interval [a, b], THEN on that interval f takes on both a maximum value and a minimum value.
EVT - Example 1 Does the function f (x) = x 2 attain a maximum or a minimum on the interval ( 1, 1)? Theorem (Extreme-Value Theorem) IF f is continuous on a bounded closed interval [a, b], THEN on that interval f takes on both a maximum value and a minimum value.
EVT - Example 2 Does the function f (x) = 1 x attain a maximum or a minimum on the interval [ 1, 1]? What about on the interval [1,2]? Theorem (Extreme-Value Theorem) IF f is continuous on a bounded closed interval [a, b], THEN on that interval f takes on both a maximum value and a minimum value.
EVT - hypotheses In the Extreme-value theorem, all the hypotheses are needed. If the interval is not bounded: The function f (x) = x 3 is continuous but has no maximum value on [0, ). If the interval is not closed: The function f (x) = x 2 is continuous but has no maximum value on the interval ( 1, 1). If the function is not continuous: The function f (x) = 1 x does not attain a maximum of a minimum on the closed and bounded interval [0, 1].
IVT + EVT Remark From the IVT, we see that continuous functions map intervals to intervals. From the Extreme-value theorem, we see that continuous functions map closed and bounded intervals to closed and bounded intervals.
Secant and Tangent Lines What is a secant of a circle? What is a tangent line to a circle?
Secant and Tangent Lines What does it mean to say that a line is tangent to a curve at at point P?
Secant and Tangent Lines Definition (Secant Line) Let f be a function and a < b be two real numbers. The unique straight line passing though the points (a, f (a)) and (b, f (b)) is called the secant line from a to b. f (b) f (a) a b The slope of this secant line is given by m ab = f (b) f (a). b a
Derivative Definition (Derivative) Let f be a function and a be a number in the domain of f. We say that f is differentiable at a if f (x) f (a) lim x a x a exists. If this limit exists, it is called the derivative of f at a and is denoted by f (a). If we let h = x a, then x a iff h 0. Therefore f f (x) f (a) f (a + h) f (a) (a) = lim = lim. x a x a h 0 h
Example 1 Example Find the derivative of the function f (x) = x 2 at x = a. Definition if the limit exists. f f (x) f (a) f (a + h) f (a) (a) = lim = lim. x a x a h 0 h
Example 2 Example Find an equation of the tangent line to the parabola y = x 2 at the point (2, 4). Definition if the limit exists. f f (x) f (a) f (a + h) f (a) (a) = lim = lim. x a x a h 0 h
The Derivative as a Function We can view the derivative as a function, with f f (x + h) f (x) f (y) f (x) (x) = lim = lim. h 0 h y x y x We say f is the derivative of f. The domain of f is the set {x : f (x) exists}.
Example 3 Example Consider the function f (x) = x 3 x. (a) Find a formula for f (x). (b) Compare the graphs of f and f. Definition if the limit exists. f f (y) f (x) f (x + h) f (x) (x) = lim = lim. y x y x h 0 h
Example 3 f (x) = x 3 x sdf f (x) = 3x 2 1
Example 3 f (x) = x 3 x sdf f (x) = 3x 2 1
Example 3 f (x) = x 3 x sdf f (x) = 3x 2 1