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C hapter 3. Limits and Their Properties. Section 3.1. A Preview of Calculus. What is Calculus. Calculus is the mathematics of change---velocities and accelerations

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## C hapter 3

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### Chapter 3

Limits and Their Properties

### Section 3.1

A Preview of Calculus

What is Calculus
• Calculus is the mathematics of change---velocities and accelerations
• Calculus is also the mathematics of tangent lines, slopes, areas, volumes, arc lengths, centroids, curvatures, and a variety of other concepts that have enabled scientists, engineers, and economists to model real-life situations
The Difference Between Precalculus and Calculus
• Precalculus deals with velocities, acceleration, tangent lines, slopes, and so on, there is a fundamental difference between precalculus.
• Precalculus is more statics, whereas calculus is more dynamic
Precalculus vs. Calculus

Precalculus

Calculus

• An object traveling at a constant velocity can be analyzed with precalculus
• The slope of a line can be analyzed with precalculus
• A tangent line to a circle can be analyzed with precalculus
• The area of a rectangle can be analyzed with precalculus
• To analyze the velocity of an accelerating object, you need calculus
• To analyze the slope of a curve you need calculus
• To analyze a tangent line to a general graph, you need calculus
• To analyze the area under a general curve, you need calculus
The Tangent Line Problem

Given a function f and a point P on its graph, you are asked to find an equation of the tangent line to the graph at point P

Excluding a vertical tangent line, the problem is equivalent to finding the slope of the tangent line at P

You can approximate this slope by using a line through the point of tangency and a second point on the curve. Such a line is called a secant line

The Tangent Line Problem Cont’d

If P(c, f(c)) is the point of tangency and Q(c + ∆x, f(c + ∆x) is a second point on the graph of f

The slope of the secant line through these two points is

msec = (f(c + ∆x) - f(c)) / (c + ∆x – c) or

msec = (f(c + ∆x) - f(c)) / (∆x)

The Tangent Line Problem Cont’d

As point Q approaches point P, the slope of the secant line approaches the slope of the tangent line. When such a “limiting position” exists, the slope of the tangent line is said to be the limit of the slope of the secant line.

EXAMPLE

The following points lie on the graph of f(x) = x2

Q1(1.5, f(1.5), Q2(1.1,f(1.1), Q3,(1.01,f(1.01),Q4(1.001, f(1.001) and Q5(1.0001, f(1.0001)

Estimate the slope of the tangent line of f at point P

The Area Problem

Find the area of a plane region bounded by the graphs of functions. This problem can be solved with a limit process. In this case, the limit process is applied to the area of a rectangle to find the area of a general region.

Consider the region bounded by the graph of the function y = f(x), the x-axis, and the vertical lines x = a and x= b

You can approximate the area of the region with several rectangular regions, the more rectangles, the better the approximation

Your goal is to determine the limit of the sum of the areas of the rectangles as the number of rectangles increase without bound.

The Area Problem - Example

Consider the region bounded by the graphs of f(x) = x2, y = 0, and x = 1

First inscribe a set of rectangles, then circumscribe a set of rectangles.

Find the sum of the areas of each set of rectangles

Use your results to approximate the area of the region

### Section 3.2

Finding Limits Graphically and Numerically

An Introduction to Limits

Let f(x) = (x3 – 1)/(x – 1), however x ≠ 1

Since you don’t know the behavior of the graph at x = 1, use two sets of values, one set that approaches 1 from the left and one set that approaches 1 from the right. Make a table.

The graph of f is a parabola with a gap

Although x cannot equal 1, you can move x arbitrarily close to 1, and as a result f(x) moves arbitrarily close to ????

An Introduction to Limits – Cont’d

Using limit notation you can write:

lim f(x) = 3

x→1

Therefore, if f(x) becomes arbitrarily close to a single L as x approaches c from either side, the limit of f(x), as x approaches c, is L and is written

lim f(x) = L

x→c

Example

Estimate the limit of each function

• f(x) = x/((x + 1)½ - 1) at 0
• f(x) = 1/x2 at 0
• f(x) = x2 at 2

### Section 3.3

Evaluating Limits Analytically

Some Basic Limits

Let b and c be real numbers and let n be a positive integer

lim b = b

x→c

lim x = c

x→c

limxn = cn

x→c

Properties of Limits

Let b and c be real numbers and let n be a positive integer, and let f and g be function with the following limits

lim f(x) = L andlim g(x) = K

x→cx→c

• Scalar multiple: lim [bf(x) = bL

x→c

• Sum or difference: lim [f(x) ± g(x)] = L ± K

x→c

• Product: lim [f(x) g(x)] = LK

x→c

• Quotient: lim [f(x)/g(x)] = L/K provided K ≠ 0

x→c

• Power: lim [f(x) n)] = Ln

x→c

Example

Find the limit of each polynomial

• lim (4x2 + 3)

x→2

• lim (x2 + x + 2)

x→1

Limits of Polynomial and Rational Functions

If p is a polynomial function and c is a real number, then

lim p(x) = p(c)

x→c

If r is a rational function given by r(x) = p(x)/q(x) and c is a real number such that q(c) ≠ 0, then

lim r(x) = r(c) = p(c)/q(c)

x→c

Example

Find the limit of the polynomial

lim (x2 + x + 2)/(x + 1)

x→1

The Limit of a Function Involving a Radical

Let n be a positive integer. The following limit is valid for c if n is odd and is valid for c>0 if n is even.

limn x = n c

x→c

The Limit of a Composite Function

If f and g are functions such that lim g(x) = L and lim f(x) = f(L), x→c

x→L

then lim f(g(x)) = f(lim g(x) = f(L)

x→cx→c

Example

Find the limit each function

lim (x2 + 4)½

x→0

lim (2x2 -10)3/2

x→3

Functions That Agree at All But One Point

Let c be a real number and let f(x) = g(x) for all x ≠ c in an open interval containing c. If the lim of g(x) as x approaches c exists, then the limit of f(x) also exists and

then lim f(x) = lim g(x)

x→cx→c

Example

Find the limit:

lim (x3 - 1)/(x -1) Hint: factor

x→1

lim [(x + 1)½ - 1]/x Hint: rationalize

x→0

The Squeeze Theorem

If h(x) ≤ f(x) ≤ g(x) for all x in an open interval containing c, except possibly at c itself, and if

lim h(x) = L = lim g(x)

x→cx→c

then lim f(x) exists and is equal to L.

x→c

### Section 3.4

Continuity and One-Sided Limits

Definition of Continuity

Continuity at a Point: A function fis continuous at cif the following three conditions are met.

• f(c) is defined
• lim f(x) exists

x→c

• lim f(x) = f(c)

x→c

Continuity on an Open Interval: A function is continuous on an open interval(a,b) if it is continuous at each point in the interval. A function that is continuous on the entire real line (-∞, ∞) is everywhere continuous.

Discontinuity

Nonremovable

Removable

• A discontinuity at c is called nonremovable if f cannot be made continuous by appropriately defining (or redefining f(c)
• Example f(x) = 1/x
• A discontinuity at c is called removable if f can be made continuous by appropriately defining (or redefining f(c)
• Example f(x) = (x2 -1)/(x -1)
One-Sided Limits and Continuity on a Closed Interval

The limit from the right means that x approaches c from values greater than c

lim f(x) = L

x→c +

The limit from the left means that x approaches c from values less than c

lim f(x) = L

x→c -

One-sided limits are useful in taking limits of functions involving radicals. For instance, if n is an even integer,

limn x = 0

x→0 +

The Existence of a Limit

Let f be a function and let c and L be real numbers. The limit of f(x) as x approaches c is L if and only if

lim f(x) = L andlim f(x) = L

x→c - x→c +

Definition of Continuity on a Closed Interval

A function f is continuous on the closed interval [a,b] if it is continuous on the open interval (a,b) and

lim f(x) = f(a) andlim f(x) = f(b)

x→a + x→b-

The function f is continuous from the right at aand continuous from the left atb

Example

Is the function f(x) = (1-x2)½continuous?

First, determine the domain of f

Next use the definition of Continuity on a Closed Interval

Properties of Continuity

If b is a real number and f and g are continuous at x = c, then the following functions are also continuous at c.

• Scalar multiple: bf
• Sum and difference: f ± g
• Product: fg
• Quotient: f/g, if g(c) ≠ 0
Types of Continuous Function at Every Point in their Domain

If b is a real number and f and g are continuous at x = c, then the following functions are also continuous at c.

• Polynomial functions: p(x) = anxn + an-1xn-1 +….
• Rational functions: r(x) = p(x)/q(x), q(x) ≠ 0
• Radical functions: f(x) = n x
Continuity of a Composite Function

If g is continuous at c and f is continuous at g(c), then the composite function given by (f ° g)(x) = f(g(x)) is continuous at c.

Intermediate Value Theorem

If f is continuous on the closed interval [a,b] and k is any number between f(a), and f(b), then there is at least one number c in[a,b]such that f(c) = k

Example

Use the intermediate value theorem to show the polynomial has a zero in the interval [0,1]

f(x) = x3 + 2x - 1

• First, is the function continuous on the interval
• Find f(0) and f(1) then compare relationship to determine if there is a zero in the interval or not

### Section 3.5

Infinite Limits

Definition of Vertical Asymptote

If f(x) approaches infinity ( or negative infinity) as x approaches c from the right or the left, then the line x =c is a vertical asymptote of the graph of f

Vertical Asymptote

Let f and g be continuous on an open interval containing c. If f(c) ≠ 0, g(c) = 0 and there exists an open interval containing c such that g(x) ≠ 0 for all x ≠ c in the interval, then the graph of the function given by

h(x) = f(x)/g(x)

Has a vertical asymptote at x = c

Example

Find the vertical asymptotes for each function

f(x) = 1/[2(x+1)]

f(x) = (x2 +1)/(x2 – 1)

f(x) = (x2 – 1)/(x – 2)

f(x) = (x2 + 2x – 8)/ (x2 – 4)

Properties of Infinite Limits

Let c and L be real numbers and let f and g be functions such that

lim f(x) = ∞ andlim g(x) = L

x→cx→c

• Sum or difference: lim[f(x) ± g(x)] = ∞

x→c

• Product: lim[f(x) g(x)] = ∞, L > 0

x→c

lim[f(x) g(x)] = -∞, L < 0

x→c

• Quotient: lim g(x)/f(x) = 0

x→c

Example

Find the limit of each function

lim (1 + 1/x2)

x→0

lim (x2 + 1)/1/(x-1))

x→1-

END