Precise Definition of the Limit

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Before we talk about what the definition is actually telling us, let’s state the precise definition of the limit, also called the Epsilon-Delta definition of the limit.

Precise Definition of the Limit

If $f(x)$ is a function for which we’re asked to find the limit as $x$ approaches $a$ , then

$\underset{x\to a}{\mathop{\lim}}f(x)=L$

if, for every $\epsilon >0$ there is some $\delta >0$ such that

$|f(x)-L|<\epsilon$ whenever $0<|x-a|<\delta$

Don’t get too caught up in the definition itself. Instead, think about it this way. Someone is going to give you some value for epsilon, $\epsilon$ . This is the distance you’re allowed to move away from the limit of the function, $L$ , along the $y$ -axis. Given the value of $\epsilon$ that they gave you, they want you to give them the value of delta, $\delta$ , that will keep you inside the epsilon range.

So if you look at the graph below, you can see that you can move delta distance away from a, in either direction along the $x$ -axis, and still fall within the epsilon range.

That’s all this means. Now, in order to prove the limit of a function using the definition, we just need to plug the information we’re given about our function and its limit into the two inequalities from the definition.

For example, if you were asked to prove that $\underset{x\to -2}{\mathop{\lim}}x^2=4$ using the Epsilon-Delta, or precise definition of the limit, you would use the following steps to prove the limit.

Steps to Prove the Limit

For the first inequality from the precise definition of the limit, $0<|x-a|<\delta$ , enter for $a$ the value that your function is approaching. For example, since we’re asked here to find the limit of the function as $x$ approaches $-2$ , enter $-2$ for $a$ , to get $0<|x+2|<\delta$ .
Leave all other parts of the inequality alone, and move onto the second inequality, $|f(x)-L|<\epsilon$ . Plug the function for which you’re asked to find the limit in for $f(x)$ , and plug the limit of the function at that point in for $L$ . In this particular example, you’d plug $x^2$ in for $f(x)$ , and you’d plug $4$ in for $L$ .
Simplify the inequality until you get one side of it to equal the value you had in the middle of the other inequality. You need to get these two values to equal one another so that you can draw a conclusion about the relationship between $\delta$ and $\epsilon$ .

Let’s work through an example in more detail.

Use the Epsilon-Delta definition of the limit to prove that $\underset{x\to 4}{\mathop{\lim}}5x+6=26$ .

Start with the first inequality, $0<|x-a|<\delta$ , and plug $4$ in for $a$ , since that is the value at which we’re evaluating the limit. For the first inequality, we therefore get $0<|x-4|<\delta$ .

Move on to the second inequality from the precise definition of the limit. We need to get one side of the second inequality to be equal to the middle of the first inequality, or in this case, $|x-4|$ . Plug the function, $5x+6$ , in for $f(x)$ , and plug the limit, 26, in for $L$ . You’ll get $|(5x+6)-26|<\epsilon$ . Simplify this inequality as much as possible, and try to get the left-hand side to equal $|x-4|$ .

$|(5x+6)-26|<\epsilon$

$|5x-20|<\epsilon$

$|5(x-4)|<\epsilon$

$|5||x-4|<\epsilon$

$|x-4|<\frac{\epsilon}{5}$

Now that the left-hand side of this second inequality is equal to the middle of the first inequality, (they are both equal to $|x-4|$ ), you can describe a relationship between $\delta$ and $\epsilon$ . Since you isolated $|x-4|$ in both inequalities, you now conclude:

If $\epsilon >0$ , then $\delta <\frac{\epsilon}{5}$ .

This means that, if you’re given any value for epsilon, you can move as far away from $x=a$ as you want, as long as the distance you move, $\delta$ , is $\delta <\frac{\epsilon}{5}$ , and you’ll still fall within the range $L-\epsilon$ to $L+\epsilon$ .

Precise Definition of the Limit

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Precise Definition of the Limit

Steps to Prove the Limit

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