Introduction to calculus
In calculus, limits are used to define continuity, derivatives and integrals. This page introduces limits and continuity, while later pages explain derivatives and integration.
In brief, it covers the following:
- How to work out if a function has a limit. If it does, how to find its value
- How to work out if a function is continuous. If it is not, how to find the values where it is not continuous
Limits
Suppose we have a function \(f(x)\). We want to look at the value this function gets close to as the input value \(x\) gets close to a particular value. We will call this value \(a.\)
In this case we say we are finding the limit of \(f(x)\) as \(x\) approaches \(a,\) and we use the following notation to represent this:
\[\lim_{x \to a} f(x)\]Sometimes we need to show whether we are approaching \(a\) from the left or from the right. This is becasue the limit does not always give the same value from each side. When approaching from the left, we use the notation \(\lim_{x \to a^-} f(x)\). This is called a left-hand limit. When approaching from the right, we use the notation \(\lim_{x \to a^+} f(x)\). This is called a right-hand limit.
For \(\lim_{x \to a} f(x)\) to exist, \(\lim_{x \to a^-} f(x)\) and \(\lim_{x \to a^+} f(x)\) need to be the same as eeach other.
We can use tables of values to calculate the left hand limit and the right hand limit of a function. This lets us work out if the limit of that function exists as \(x\) approaches a given value.
Example 1:
Use tables of values to investigate \(\lim_{x \to 3} (x^2 + 4x + 5)\)
Solution:
Limit as \(x\) approaches \(3\) from the left:
| \(x\) | \(x^2 + 4x + 5\) |
|---|---|
| 2.9 | 25.01 |
| 2.99 | 25.9001 |
| 2.999 | 25.990001 |
| 2.9999 | 25.99900001 |
Limit as \(x\) approaches \(3\) from the right:
| \(x\) | \(x^2 + 4x + 5\) |
|---|---|
| 3.1 | 27.01 |
| 3.01 | 26.1001 |
| 3.001 | 26.010001 |
| 3.0001 | 26.00100001 |
These tables indicate that:
\(\lim_{x \to 3^-} (x^2 + 4x + 5) =\) \(\lim_{x \to 3^+} (x^2 + 4x + 5) =\) \(26\)
Therefore:
\(\lim_{x \to 3} (x^2 + 4x + 5) = 26\)
The fact that \(\lim_{x \to 3} (x^2 + 4x + 5) = 26\) should not be surprising, as this is the value of \(x^2 + 4x + 5\) when \(x = 3\). In other words, \(\lim_{x \to 3} (x^2 + 4x + 5) = f(3)\)
In fact, the following is true for any function that is continuous at \(x = a\):
\(\lim_{x \to a} f(x) = f(a)\)
More information on continuity, and the conditions required for a function to be continuous at \(x = a,\) are covered in the following section.
Example 2:
Use tables of values to investigate \(\lim_{x \to 2} (\frac{1}{x-2})\)
Solution:
Limit as \(x\) approaches \(2\) from the left:
| \(x\) | \(x^2 + 4x + 5\) |
|---|---|
| 1.9 | -10 |
| 1.99 | -100 |
| 1.999 | -1000 |
| 1.9999 | -10000 |
| 1.99999 | -100000 |
Limit as \(x\) approaches \(2\) from the right:
| \(x\) | \(x^2 + 4x + 5\) |
|---|---|
| 2.1 | 10 |
| 2.01 | 100 |
| 2.001 | 1000 |
| 2.0001 | 10000 |
| 2.00001 | 100000 |
These tables indicate that:
\(\lim_{x \to 2^-} (\frac{1}{x-2}) = -\infty\)
\(\lim_{x \to 2^+} (\frac{1}{x-2}) = \infty\)
Therefore:
\(\lim_{x \to 2} (\frac{1}{x-2})\) does not exist
It is important to understand something about infinity here. Technically, a limit cannot equal positive infinity or negative infinity. This means that if a limit tends towards positive or negative infinity, the limit does not exist.
However, saying that the limit tends towards positive or negative infinity (as shown above) still gives us useful information. It tells us more about how the function behaves close to the specified point.
The fact that this limit does not exist also tells us something else. It means that this function is not continuous. More information about continuity is provided in the following section.
For now, test your understanding of limits by working through the questions below.
Using tables of values, or by substituting \(x = 3\) into the function (since it is continuous at this point), we can determine that \(\lim_{x \to 3} (x^2 + 4) = 13\)
Using tables of values, we can determine that \(\lim_{x \to 1} (\frac{x^2-3x+2}{x-1}) = -1\)
Using tables of values, we can determine that this limit does not exist. If we want to provide more information about the limit, however, we could state that:
- \(\lim_{y \to -5^-} (\frac{y^2+4}{y+5}) = -\infty\)
- \(\lim_{y \to -5^+} (\frac{y^2+4}{y+5}) = \infty\)
Using tables of values, or by substituting \(x = 2\) into \(f(x) = 5 - x^2\), we can determine that:
\(\lim_{x \to 2^-} f(x) =1\)
Similarly, using tables of values or by substituting \(x = 2\) into \(f(x) = x - 1\), we can determine that:
\(\lim_{x \to 2^+} f(x) =1\)
Therefore, \(\lim_{x \to 2} f(x) =1\)
Continuity
A function is called continuous if its graph has no breaks or jumps. This means you could draw the whole graph without ever lifting your pencil off the paper. If a function is continuous everywhere on its graph, we say it is continuous.
Sometimes, however, a function is not continuous everywhere. In many cases, the function might have just one point where it is not continuous. This means it is continuous at every other value of \(x\), except for that one point.
For example:
The idea above is a helpful way to think about continuity, but there is also a formal mathematical definition. This definition says that a function \(f(x)\) is continuous at \(x = a\) only if all of the following are true:
- The function must have a value (or exist) at \(x = a\).
- The limit of \(f(x)\) as \(x\) approaches \(a\) must exist. This means the limit from the left side, \(\lim_{x \to a^-}f(x)\), must equal the limit from the right side, \(\lim_{x \to a^+}f(x)\). This looks like \(\lim_{x \to a^-}f(x) = \lim_{x \to a^+}f(x)\)
- The limit of \(f(x)\) as \(x\) approaches \(a\) must be equal to the value of the function at \(a\), which is \(f(a)\). This looks like \(\lim_{x \to a}f(x) = f(a)\)
Example:
Is the function \(f(x) = \frac{x+4}{x^2 + 3x - 40}\) continuous? If not, for which value or values is it discontinuous?
Solution:
Noting that dividing by \(0\) is undefined, we can factorise the denominator of this function to find the values of \(x\) such that it is equal to \(0.\) Factorising gives the following:
\(f(x) = \frac{x+4}{(x+8)(x-5)}\)
Finding the values of \(x\) such that \((x+8)(x-5) = 0\) gives \(x = -8\) and \(x = 5,\) which means the function is discontinuous at these values.
All polynomials are continuous everywhere. This means that if you have two functions \(f(x)\) and \(g(x)\), and both of them are continuous at \(x = a\) then the following are also true:
- \(k f(x)\) is continuous at \(x = a\) (where \(k\) is any real number)
- \(f(x) + g(x)\) is continuous at \(x = a\)
- \(f(x) - g(x)\) is continuous at \(x = a\)
- \(f(x) \times g(x)\) is continuous at \(x = a\)
- \(f(x) \div g(x)\) is continuous at \(x = a\), as long as \(g(a) \neq 0\)
Test your understanding of continuity by working through the questions below:
Yes, this function is continuous everywhere as it is a polynomial.
The denominator of this function is equal to \(0\) when \(x = 3,\) hence the function is discontinuous when \(x = 3\)
Factorising the denominator of this function gives:
\(f(x) = \frac{x - 2}{(x+8)(x+2)}\)
Since you can’t divide by \(0\) we need to find the values for \(x\) such that \((x+8)(x+2) = 0,\) which gives \(x = -8\) and \(x = -2\)
This means the function is discontinuous at these values.
\(h(x) = f(x) + g(x),\) and since \(f(x)\) and \(g(x)\) are both continuous at \(x = 1\) it follows that \(h(x)\) must also be continuous at \(x = 1\)