Continuity

When we discuss the idea of limits, we often discuss functions that are undefined at specific points. These functions typically require us to evaluate limits as we approach the undefined areas. From this idea, limits can tell us another thing about a function, which is the continuity of a function.

Definition: A function f is continuous at a number a, if lim_{x \to a}f(x) = f(a)

This requires a few things, for one, f(a) has to exist and be defined. In addition to this, lim_{x \to a} f(x) must exist as well. If a function is not continuous at a number a, it is considered to be discontinuous at a.

When we looked at functions such as \frac{x^2-1}{x-1}, we saw that we could factor and remove the undefined area, leaving us with just x+1. When we can remove a discontinuity, we refer to it as removable. Just as we could consider limits from the left and right of a number, we can also consider continuity from the left and right of a number.

Definition: A function f is continuous from the right at a if lim_{x \to a^+}f(x) = f(a). Similarly, a function f is continous from the left of a if lim_{x \to a^-}f(x) = f(a).

Continuity is an important concept as it tells us a lot about the behavior of a function. Knowing where a function is or is not defined is valuable information.

Definition: A function f is continuous on an interval if it is continuous at every number in the interval.

Of course, it makes sense to want to know continuity at a range of points rather than just a single point. Using this definition, we can show that a function is continuous on any interval.

Example: Show that the function f(x) = 1 - \sqrt{1-x^2} is continous on the interval [-1,1]

To do this, we first pick a variable to be any number that is in the range defined. Let a be a number, such that -1 < a < 1, or a is in the interval [-1,1]. We can complete the following steps to solve the limit:
lim_{x \to a} 1 - \sqrt{1-x^2}
= lim_{x \to a} 1 - \sqrt{lim_{x \to a} 1 - x^2}
= 1 - \sqrt{1-a^2}

The function = 1 - \sqrt{1-a^2} is exactly the function f(x), if we evaluated it at a. Due to this, we can conclude that lim_{x \to a} 1 - \sqrt{1-x^2} = f(a), therefore the function is continous on the interval [-1,1]

An interesting result comes from the limit laws we discussed in the previous section. If we apply the limit laws, we can make some assumptions about continuity as well.

Theorem: If f and g are continuous, and c is constant, the following are also constant:

  1. f(x)+g(x)
  2. f(x)-g(x)
  3. c*f(x)
  4. f(x)*g(x)
  5. \frac{f(x)}{g(x)}, provided g(x) \ne 0

To prove any of these theories, we simply need to follow through with the limit laws. For example, consider the first conclusion, that f+g is also continuous. From limit laws, we know:
lim_{x \to a} (f(x) + g(x)) = lim_{x \to a}f(x) + lim_{x \to a}g(x).
If f(x) and g(x) are continous at a, it means that their limits are equal to f(a) and g(a) respectively. Therefore:
lim_{x \to a} (f(x) + g(x)) = lim_{x \to a}f(x) + lim_{x \to a}g(x) = f(a) + g(a)
which is continous.

It is good to try the other conclusions are exercises to see if you can understand the argument being presented. Proofs are an important part of math, and the sooner you try them, the better you will get at them.

Following the limit laws discussed in the last section, we can derive some information about polynomials and rational functions.

Theorem: Any polynomial is continuous for all real numbers. Any rational function is continuous on its domain.

Let’s first consider polynomial functions. We know that a polynomial function is one that is in the form:
f(x) = c_n*x^n + c_{n-1} * x^{n-1}+...+c_0

If we apply a limit approaching some number a, which is a real number, we can apply limit laws as follows:
lim_{x \to a} c_n*x^n + c_{n-1} * x^{n-1}+...+c_0
=lim_{x \to a} c_n*x^n + lim_{x \to a} c_{n-1} * x^{n-1}+...+  lim_{x \to a}c_0
=c_n * lim_{x \to a} x^n +   c_{n-1} * lim_{x \to a} x^{n-1}+...+c_0

Since all of the x terms are power terms, we can apply the power law for limits to get:
c_n*a + c_{n-1}*a+...+c_0 = f(a)

Therefore, polynomials are continous for all real numbers.

To see how this applies for rational functions, we can apply a similar logic. We know that a rational function is defined as: f(x) = \frac{g(x)}{h(x)}, where g(x) and h(x) are polynomials.

For some a that is in the real numbers, where a is such that h(x) != 0, we have:
lim_{x \to a} \frac{g(x)}{h(x)} = \frac{lim_{x \to a} g(x)}{lim_{x \to a} h(x)}

We just proved that for polynomials, lim_{x \to a} f(x) = f(a). Therefore:
\frac{lim_{x \to a} g(x)}{lim_{x \to a} h(x)} = \frac{g(a)}{h(a)} = f(a)

Therefore, the theory stands.  

The reason that continuity is important is that it makes solving limits easier, assuming we can show a function is continuous.

Example: Find lim_{x \to 2} \frac{x^2+3}{x^2}

Since we know that all rational functions are continuous, we can simply evaluate this at 2 in order to find the limit.

lim_{x \to 2} \frac{x^2+3}{x^2} = \frac{2^2+3}{2^2} = \frac{7}{4}

There are many other functions that we know are continuous.

Theorem: The following functions are continuous on their domain:

  1. Polynomials
  2. Rationals
  3. Root Functions
  4. Trig Functions
  5. Inverse trig functions
  6. Exponential functions
  7. Logarithmic functions

So, if we ever see any of these functions in a limit, and the value is in the domain, we can simply sub the limit value in, and solve the function.

There is one other result of continuity that is important at this point in Calculus, and that is the Intermediate Value Theorem.

Intermediate Value Theorem: Suppose f is continuous on the closed interval [a,b]. If N is a number between f(a) and f(b), where f(a) \ne f(b), then there must exist a number c in (a,b), such that f(c) = N

Why is this true? To understand this theorem, consider the opposite of this theorem. Suppose that f is continuous on a closed interval, [a,b], and N is between f(a) and f(b), f(a) \ne f(b). Now, let’s assume that there is no c in (a,b) such that f(c) = N.

If this statement is true, it means that somewhere in (a,b), there must be an undefined portion. This is because if a function is continuous between [a,b], it must reach every value between a and b. If there is an undefined portion, it means that f is not continuous on [a,b], which is a contradiction. Therefore, it must be true that there is a c in (a,b) such that f(c) = N.


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