In this lesson, we will look at the concept of the limit of a function. Limits enable us to evaluate the value that a function approaches for a given input. This will be very useful in enabling us to further develop our understanding of the behaviour of functions.
Let's look at some examples to learn more about the concept of a limit.
Exploration
Let's consider the limit of the function $f(x)=x^2$f(x)=x2 as $x$x approaches $3$3.
This is written as follows:
$\lim_{x\rightarrow3}\ x^2$limx→3 x2.
The value that the input is approaching is placed under the limit symbol. The function is placed to the right of the limit symbol. We can consider what this means graphically.
We will sketch the function and create a table of values close to the input of our limit.
| $x$x | $f(x)=x^2$f(x)=x2 |
|---|---|
| $2.9$2.9 | $8.41$8.41 |
| $2.99$2.99 | $8.9401$8.9401 |
| $2.999$2.999 | $8.994001$8.994001 |
| $3.001$3.001 | $9.006001$9.006001 |
| $3.01$3.01 | $9.0601$9.0601 |
| $3.1$3.1 | $9.61$9.61 |
From the graph and table above, we can observe that as $x$x approaches $3$3, the function $f(x)=x^2$f(x)=x2approaches $9$9. Approaching the value $x=3$x=3 from either direction results in the same limiting value of $9$9, which is also the value of the function at $x=3$x=3. For many functions such as this, the limit at a particular point is simply the function evaluated at that point. We can solve limit problems algebraically and the next few examples will demonstrate this.
Worked examples
Example 1
Evaluate the limit of $f(x)=x^2+2x-1$f(x)=x2+2x−1 as $x$x approaches $-2$−2.
Think: This can be written as the limit, $L=\lim_{x\rightarrow-2}x^2+2x-1$L=limx→−2x2+2x−1. We can solve this by substituting in $x=-2$x=−2
Do: Making the substitution the solution is $L=(-2)^2+2(-2)-1=-1$L=(−2)2+2(−2)−1=−1.
Example 2
Evaluate $\lim_{x\rightarrow2}\ \frac{4x^2-8x}{x-2}$limx→2 4x2−8xx−2.
Think: The function is not defined for $x=2$x=2 as there is $x-2$x−2 in the denominator and when $x=2$x=2 the denominator would equal $0$0 which is not defined. Hence, we cannot simply substitute in the value. Rather than looking at a graph or table we can simplify this function by factorising the top and dividing by the numerator.
Do: Simplifying the function:
| $f(x)$f(x) | $=$= | $\frac{4x^2-8x}{x-2}$4x2−8xx−2 |
| $f(x)$f(x) | $=$= | $\frac{4x\left(x-2\right)}{x-2}$4x(x−2)x−2 |
| $f(x)$f(x) | $=$= | $4x$4x, for $x\ne2$x≠2 |
Thus for this function, everywhere except $x=2$x=2, the function will have the value $4x$4x. This means approaching from either side will result in approaching $4x$4x evaluated at $x=2$x=2, which is the limit $8$8. We can write:
| $\lim_{x\rightarrow2}\frac{4x^2-8x}{x-2}$limx→24x2−8xx−2 | $=$= | $\lim_{x\rightarrow2}\frac{4x\left(x-2\right)}{x-2}$limx→24x(x−2)x−2 |
| $=$= | $\lim_{x\rightarrow2}\ 4x$limx→2 4x | |
| $=$= | $8$8 |
Notice the $\lim_{x\rightarrow2}$limx→2 remains in the expression until the limit is "taken" by substituting in the value $2$2. This technique of simplifying a rational expression to reveal the behaviour of the graph, excluding the point of discontinuity, is very important and we will make wide use of it in calculus.
The graph of this function will look like the graph of $y=4x$y=4x but with a single missing point at $x=2$x=2. As seen below.
The definition of a limit
A function $f(x)$f(x) has a limit $L$L at $x=a$x=a if the function gets closer to $L$L as $x$x gets closer and closer to $a$a. The notation for this is:
$\lim_{x\rightarrow a}f(x)=L$limx→af(x)=L
This can be read as "the limit of $f(x)$f(x) as $x$x approaches $a$a is $L$L" or "as $x$x approaches $a$a, $f(x)$f(x) approaches $L$L".
Practice questions
Question 1
Consider the function $f\left(x\right)=5x^2+1$f(x)=5x2+1.
Complete the table to find the exact values of $f\left(x\right)$f(x) as $x$x gets closer and closer to $2$2 from the left, and closer and closer to $2$2 from the right.
$x$x $1.9$1.9 $1.99$1.99 $1.999$1.999 $2.001$2.001 $2.01$2.01 $2.1$2.1 $f\left(x\right)$f(x) $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Hence, find the value of $\lim_{x\to2}\left(5x^2+1\right)$limx→2(5x2+1).
Question 2
Use limit notation to describe the following sentence:
The value that the function $y=x+4$y=x+4 approaches as $x$x approaches $168$168.
$\lim_{\editable{}\to\editable{}}\left(\editable{}\right)=\editable{}$lim→()=
question 3
Find $\lim_{x\to-3}\left(\frac{x^2+3x}{x^2-9}\right)$limx→−3(x2+3xx2−9).