Calculus 2.5 Continuity

Name: Calculus 2.5 Continuity
Uploaded: 2020-08-06T16:00:03.000Z
Duration: 1 h 20 min 44 s
Description: My notes are available at http://asherbroberts.com/ (so you can write along with me). Calculus: Early Transcendentals 8th Edition by James Stewart

Continuity of Functions

In this section, we learn about the concept of continuity for functions. A function is said to be continuous at a number if the limit of the function as x approaches that number is equal to the value of the function at that number. We explore the conditions that need to be met for a function to be continuous.

Definition of Continuity

A function f is continuous at a number a if:

The limit of f(x) as x approaches a exists.

The value of f(a) exists.

The limit and value are equal.

Identifying Discontinuities

Discontinuities occur when any of the three conditions for continuity are not met.

Discontinuities can be identified by analyzing the graph of the function.

Example 1: Discontinuity at x = 1

At x = 1, there is a hole in the graph, indicating that the function is not defined at that point.

Since f is not defined, it implies that f is discontinuous at x = 1.

Example 2: Discontinuity at x = 3

At x = 3, although the function is defined, the limit from both sides does not match with the value of the function.

The limit as x approaches 3 does not exist, indicating discontinuity at x = 3.

Example 3: Discontinuity at x = 5

At x = 5, both limits from left and right exist and are equal.

However, neither limit matches with the value of the function at x = 5.

Therefore, there is discontinuity at x = 5.

Analyzing Specific Functions

Function: f(x) = (x^2 - x - 2) / (x - 2)

The function is undefined when the denominator is zero, which occurs at x = 2.

Therefore, f is not defined, indicating discontinuity at x = 2.

Function: f(x) = 1 / x^2 (if x ≠ 0), f(x) = 1 (if x = 0)

Although the function is defined at x = 0, the limit as x approaches 0 does not exist.

The limit of 1/x^2 as x approaches 0 becomes infinity.

Hence, there is discontinuity at x = 0.

Function: f(x) = (x^2 - x - 2) / (x - 2) (if x ≠ 2), f(x) = 3 (if x = 2)

The function is defined for all values except at x = 2.

Analyzing the limit as x approaches to from both sides yields a value of 3.

However, the value of the function at x = 2 is equal to one.

Since the limit and value do not match, there is discontinuity at x = 2.

Function: Greatest Integer Function

The greatest integer function chops off any decimal part and returns the largest integer less than or equal to a given number.

When approaching any integer from left or right, the limits do not match with the corresponding values of the function.

Therefore, there are discontinuities for every integer value.

By understanding continuity and identifying discontinuities in functions, we can analyze their behavior and properties more effectively.

Discontinuity of a Function

In this section, the speaker discusses the concept of discontinuity in functions and how it relates to limits.

Discontinuity at Integer Values

A function f(x) does not exist as x approaches n for any integer n in the set of integers (Z).

The limit of f(x) as x approaches an integer value does not equal the value of the function.

Therefore, f(x) is discontinuous at every integer value.

Right and Left Continuity

A function is continuous from the right at a number 'a' if the limit of f(x) as x approaches 'a' from the right is equal to the value of the function.

Similarly, a function is continuous from the left at 'a' if the limit of f(x) as x approaches 'a' from the left is equal to the value of the function.

Continuity Analysis for Greatest Integer Function

The greatest integer function has potential issues only at integer values.

From approaching these values from the right, we can see that it converges to its corresponding integer value.

Therefore, it is right continuous at every integer.

However, when approaching these values from the left, it converges to one less than its corresponding integer value.

Hence, it is not left continuous and breaks continuity.

General Definition of Continuity on an Interval

A function f is continuous on an interval if it is continuous at every number within that interval.

If a function is defined only on one side of an endpoint in an interval, continuity at that endpoint means either continuity from the right or continuity from the left.

Continuity Analysis for a Specific Function

In this section, a specific function is analyzed for continuity on a given interval.

Continuity Analysis for f(x) = 1 - sqrt(1 - x^2)

The function f(x) is defined as 1 minus the square root of (1 minus x squared).

It is continuous on the open interval from -1 to 1.

To check continuity at the endpoints, we need to analyze it from both the right and left sides.

Continuity from the Right at -1

As we approach -1 from the right side, the limit of f(x) equals 1.

Therefore, it is continuous from the right at -1.

Continuity from the Left at 1

As we approach 1 from the left side, the limit of f(x) also equals 1.

Hence, it is continuous from the left at 1.

Conclusion

The function f(x) = 1 - sqrt(1 - x^2) is continuous on the entire closed interval [-1, 1].

Continuity of Functions and Operations

In this section, properties of continuity are discussed in relation to functions and operations.

Properties of Continuous Functions

If functions f and g are continuous at a point 'a', then their sum (f + g), difference (f - g), constant times function (c * f), product (f * g), and quotient (f / g) are also continuous at 'a'.

This holds true assuming that g is not zero at 'a'.

The remaining properties are not summarized as they require further explanation beyond what can be provided in bullet points.

Continuity of Polynomials and Rational Functions

In this section, the speaker discusses the continuity of polynomials and rational functions. They explain that any polynomial is continuous everywhere on the real line, and any rational function is continuous wherever it is defined.

Polynomial Continuity

A polynomial function can be written as a sum of terms, where each term is a constant multiplied by a power of x.

The limit of x to the power of m as x approaches any value 'a' is equal to 'a' raised to the power of 'm'.

This implies that x to the power of m is continuous for any value of 'm'.

Similarly, a constant function is also continuous.

Since a polynomial can be expressed as a sum or product of these continuous functions, it follows that polynomials are continuous.

Rational Function Continuity

Rational functions are quotients of polynomials.

If polynomials are continuous, then rational functions must also be continuous in their domains.

Continuity of Other Types of Functions

In this section, the speaker mentions several other types of functions that are continuous at every number in their domains. These include root functions, trigonometric functions (such as sine and cosine), inverse trigonometric functions, exponential functions, and logarithmic functions.

Example: Continuity Analysis

In this section, the speaker provides an example to analyze the continuity of a given function.

Function Analysis

The function f(x) = ln(x) + tan^(-1)(x) / (x^2 - 1) is given.

The natural logarithm function (ln(x)) is continuous for all positive numbers.

The tangent inverse function (tan^(-1)(x)) is continuous everywhere.

The denominator, x^2 - 1, is a polynomial and therefore continuous everywhere.

The potential problem areas are negative numbers (ln(x) is not defined for negative numbers) and any number that makes the denominator equal to zero.

Continuity Conclusion

Based on the analysis, the function f(x) is continuous on the intervals (0, 1) and (1, infinity), excluding 0 and 1.

Evaluating a Limit

In this section, the speaker demonstrates how continuity allows for direct substitution when evaluating limits.

Example: Evaluating a Limit

The limit of sin(x)/2 + cos(x) as x approaches pi is to be evaluated.

Since cosine can take any value between -1 and 1, the smallest possible value for the denominator (2 - 1 = 1).

As long as the denominator is never equal to zero, both sine and cosine are continuous functions at every number in their domains.

Therefore, the given function is continuous everywhere.

By continuity, if we let f(x) represent the function, then f(pi) must be equal to the limit as x approaches pi of f(x).

Substituting pi into sine gives us zero in the numerator. The value of cosine does not affect this result since it appears only in the denominator.

Composition of Continuous Functions

In this section, the speaker discusses composition of functions and how continuity allows for moving limits inside functions.

Composition of Functions

If a function f is continuous at b and g has a limit approaching a that equals b, then the limit of f(g(x)) as x approaches a will be equal to f(b).

This means that when dealing with compositions involving continuous functions, we can move limits inside functions.

Proof and Definition

The proof involves an epsilon-delta argument, but it is not as difficult as it may seem.

By the definition of continuity, the limit of f(y) as y approaches b is equal to f(b).

This means that for every epsilon, there exists a delta such that if the distance between y and b is less than delta, then the distance between f(y) and the limit is less than epsilon.

The transcript does not provide further sections or timestamps.

[t=0:27:11s] Fact 1: Continuity at b implies the limit of f(x) as x approaches b is equal to f(b)

In this section, the speaker discusses the first fact about continuity. They explain that if a function f is continuous at a point b, then the value of the function at b is equal to the limit of the function as x approaches b.

The first fact states that if a function f is continuous at a point b, then the value of the function at b is equal to the limit of f.

This fact can be proven using an epsilon-delta argument.

By definition, for any epsilon value chosen by an opponent, there exists a delta such that if |x - b| < delta, then |f(x) - f(b)| < epsilon.

The speaker explains that instead of using a general epsilon value, they use delta one to show that it works for every possible epsilon value.

By letting y = g(x), they show that if g(x) - b < delta one, then y - b = delta one for y = g(x).

If g(x) - b < epsilon, then by definition, the limit exists and equals f(b).

Therefore, they have proved that if f is continuous at b and g is continuous at x = a (where g(a) = b), then their composition f(g(x)) is also continuous.

[t=0:30:31s] Fact 2: Passing through limits in composite functions

In this section, the speaker introduces Fact 2 about passing through limits in composite functions. They explain how this fact allows them to prove continuity in composite functions.

Fact 2 states that if we have a composition of two functions where both are continuous (f composed with g), then we can pass the limit inside the function.

The speaker demonstrates this fact by taking the example of finding the limit of arcsin(1 - sqrt(x) / (1 - x)) as x approaches 1.

They explain that since arcsin is continuous, they can pass the limit inside and simplify the expression to arcsin(1 / (1 + sqrt(x))).

By plugging in x = 1, they find that the limit equals pi/6.

[t=0:33:37s] Composite functions and continuity

In this section, the speaker discusses how composite functions are continuous and provides an example to illustrate this concept.

The speaker states that composite functions are continuous based on Fact 2.

They explain that if we write functions as their composites, we can show their continuity.

To demonstrate this, they consider a function h(x) = sin(x^2) and rewrite it as f(g(x)), where g(x) = x^2 and f(x) = sin(x).

Since both g(x) and f(x) are continuous everywhere, their composition h(x) must also be continuous everywhere.

[t=0:35:21s] Discontinuity in composite functions

In this section, the speaker discusses an example to show where a composite function may be discontinuous.

The speaker takes the example of h(x) = ln(1 + cos(x)).

They explain that ln is defined when its argument is positive, so h is defined when 1 + cos(x) > 0.

However, there is a discontinuity when 1 + cos(x) equals zero because ln cannot be evaluated for negative values.

They determine that 1 + cos(x) equals zero when cos(x) equals -1 or at multiples of pi minus pi.

Therefore, h is discontinuous at those points.

The transcript is already in English, so no language conversion is needed.

Intermediate Value Theorem

In this section, the concept of the Intermediate Value Theorem is explained. The theorem states that if a function is continuous on a closed interval and takes on two different values at the endpoints of the interval, then it must also take on every value in between those two values at some point within the interval.

Explaining the Intermediate Value Theorem

The Intermediate Value Theorem states that for a function f(x) that is continuous on a closed interval [a, b], if there exists a number n between f(a) and f(b), where f(a) is not equal to f(b), then there must exist a number c in the open interval (a, b) such that f(c) is equal to n.

This means that if you have two values of your function and you have a number in between those values, then your function will be equal to that number somewhere between those values.

Applying the Intermediate Value Theorem to Find Roots

To demonstrate how to use the Intermediate Value Theorem to find roots of an equation, let's consider the equation 4x^3 - 6x^2 + 3x - 2 = 0 and find a root between 1 and 2.

We need to show that the function f(x) is below zero (negative) at one point and above zero (positive) at another point between one and two. According to the Intermediate Value Theorem, this guarantees that there exists a root within that interval.

Let's define f(x) as 4x^3 - 6x^2 + 3x - 2. We want to find a root, which means we want f(x) to be equal to zero (n = 0).

Evaluating f, we get -1, which is negative. Evaluating f, we get 12, which is positive. Therefore, the value zero lies between f and f.

Since f(x) is a polynomial function and polynomials are continuous everywhere, it is also continuous on the closed interval [1, 2]. The Intermediate Value Theorem guarantees that there exists some number c in the open interval (1, 2) such that f(c) = 0.

In other words, there is an x value that matches up to the y value of zero between any two y values of the function.

By applying the Intermediate Value Theorem, we can determine the existence of roots for a given equation within a specified interval.