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Theorems of Continuity

Q: What is the theorem of continuous values?

There isn't one. Maybe you mean the Intermediate Value Theorem?

Q: What is intermediate value theorem of continuous function?

That a continuous function on a compact set will attain its minimum and maximum values.

Q: How do you prove the continuity theorem?

Well, there is more than one theorem about continuity. Proving most of them involves using limits.

Q: What is the continuity theorem?

There are quite a few. There is one about sums, one about compositions, the Squeeze Theorem and the Intermediate Value Theorem.

Q: What are the 3 conditions of continuity?

The function is defined at the point, the limit exists at the point, and that the limit is equal to the function value at that point.

By now you are probably wondering why there are so many darn theorems in calculus. Ones about limits, ones about derivatives, ones about integrals, theorems all over the place! That is because by and large it is much easier to prove one theorem and apply it to lots of different functions than it is to prove things for each and every different kind of function. Really the number of theorems in calculus is just a reminder that mathematicians are efficient.

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Theorems of Continuity for Functions

Theorems of continuity rely heavily on what you already know about limits. For a review on limits see Limits and Finding Limits. This first theorem follows directly from the definition of continuity and the properties of limits.

Theorem: Properties of Continuous Functions

Suppose $f (x)$ and $g (x)$ are functions which are continuous at $x = p$ . Then the following are true:

Sum Property: $(f + g) (x)$ is continuous at $x = p$ ;

Difference Property: $(f - g) (x)$ is continuous at $x = p$ ;

Product Property: $(f \cdot g) (x)$ is continuous at $x = p$ ;

Constant Multiple Property: if $k$ is a real number then $(k \cdot f) (x)$ is continuous at $x = p$ ;

Quotient Property: if $g (p) \neq 0$ then $(\frac{f}{g}) (x)$ is continuous at $x = p$ .

Suppose

$f (x) = |x|$ and $g (x) = \frac{x + 3}{x - 1}$ .

Where is $(f \cdot g) (x)$ continuous?

Answer:

Remember that the absolute value function is continuous everywhere, and rational functions are continuous on their domains.

Since $g (x)$ is a rational function with a domain of $(- \infty, 1) \cup (1, \infty)$ , it is continuous everywhere except at $x = 1$ .

Then using the Product Property, $(f \cdot g) (x)$ is continuous everywhere except at $x = 1$ .

Notice that the composition of functions isn't listed in the Properties of Continuous Functions. There is a separate theorem for compositions because it is proven using domains of functions and the definition of continuity rather than by using limits like the Properties of Continuous Functions.

Theorem: Composition of Continuous Functions

If $g (x)$ is continuous at $x = p$ , and $f (x)$ is continuous at $g (p)$ , then $f (g (x)) = (f \circ g) (x)$ is continuous at $x = p$ .

If you know that $g (x)$ is continuous at $x = p$ , is $|g| (x)$ continuous at $x = p$ ?

Answer:

Take $f (x)$ to be the absolute value function, which is continuous everywhere.

That means you know that $g (x)$ is continuous at $x = p$ and $f (x)$ is continuous at $g (p)$ .

Then using the Composition of Continuous Functions theorem, $(f \circ g) (x) = |g| (x)$ is continuous at $x = p$ .

Theorems on Discontinuity

You might be wondering why there are plenty of theorems for continuous functions, and no equivalent ones for discontinuity. Let's look at an example to show why not.

Take

$f (x) = \{\begin{cases} 1, & x < 0 \\ - 1, & x \geq 0 \end{cases}$ and $g (x) = \{\begin{cases} - 1, & x < 0 \\ 1, & x \geq 0 \end{cases}$ .

Neither of these functions is continuous at $x = 0$ .

If you add these two functions which aren't continuous at $x = 0$ , is their sum still discontinuous at $x = 0$ ? Well,

$(f + g) (x) = \{\begin{cases} 1 + (- 1), & x < 0 \\ - 1 + 1, & x \geq 0 \end{cases} = \{\begin{cases} 0, & x < 0 \\ 0, & x \geq 0 \end{cases} = 0$ ,

which is continuous everywhere. So by counterexample, you have shown that there can't be a Sum Property for discontinuous functions.

What about the Product Property? Their product is

$(f \cdot g) (x) = \{\begin{cases} 1 \cdot (- 1), & x < 0 \\ - 1 \cdot 1, & x \geq 0 \end{cases} = \{\begin{cases} - 1, & x < 0 \\ - 1, & x \geq 0 \end{cases} = - 1$ ,

which is continuous everywhere. So you can't have a Product Property for functions which aren't continuous.

You might be thinking that nothing can go wrong with the Constant Property. It is just multiplying by a constant!

In fact, that one goes wrong too.

Take $k = 0$ . Then $(k \cdot f) (x) = (0 \cdot f) (x) = 0$ , which is continuous everywhere. So even the Constant Property doesn't hold for discontinuous functions.

Similar to the functions in the example above, you can come up with functions to show that the Difference Property and Quotient Property also don't hold for discontinuous functions.

Theorems of Continuity - Key takeaways

Sum Property for continuous functions: Suppose $f (x)$ and $g (x)$ are functions which are continuous at $x = p$ . Then $(f + g) (x)$ is continuous at $x = p$ .
Difference Property for continuous functions: Suppose $f (x)$ and $g (x)$ are functions which are continuous at $x = p$ . Then $(f - g) (x)$ is continuous at $x = p$ .
Product Property for continuous functions: Suppose $f (x)$ and $g (x)$ are functions which are continuous at $x = p$ . Then $(f \cdot g) (x)$ is continuous at $x = p$ .
Constant Multiple Property for continuous functions: Suppose $f (x)$ is a function which is continuous at $x = p$ . Then if $k$ is a real number, $(k \cdot f) (x)$ is continuous at $x = p$ .
Quotient Property for continuous functions: Suppose $f (x)$ and $g (x)$ are functions which are continuous at $x = p$ . Then if $g (p) \neq 0$ , $(\frac{f}{g}) (x)$ is continuous at $x = p$ .
Composition of continuous functions: If $g (x)$ is continuous at $x = p$ , and $f (x)$ is continuous at $g (p)$ , then $f (g (x)) = (f \circ g) (x)$ is continuous at $x = p$ .
None of the above properties hold in general for functions that are discontinuous.

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Frequently Asked Questions about Theorems of Continuity

What is the theorem of continuous values?

There isn't one. Maybe you mean the Intermediate Value Theorem?

What is intermediate value theorem of continuous function?

That a continuous function on a compact set will attain its minimum and maximum values.

How do you prove the continuity theorem?

Well, there is more than one theorem about continuity. Proving most of them involves using limits.

What is the continuity theorem?

There are quite a few. There is one about sums, one about compositions, the Squeeze Theorem and the Intermediate Value Theorem.

What are the 3 conditions of continuity?

The function is defined at the point, the limit exists at the point, and that the limit is equal to the function value at that point.

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