Want to know if a function has a strange jump in it? Looking at the limit of the function can give you that information.
Finding the Limit of a Function
The idea of the limit of the function having the value \(L\) at the point \(x=a\) is that if other \(x\) values are close to \(a\), then the function values at those \(x\) values are close to \(L\).
In other words, someone hands you a function and a point \(x=a\). You get to propose a limit \(L\), then they give you the \( \epsilon \) that tells you how close to \(L\) all the function values need to be. Lastly, you get to see if you can find a \(\delta\) so that if your \(x\) values are within the \(\delta\) window you are guaranteed that all of the function values will be within their \(\epsilon\) window of \(L\). So let's look at some examples.
The function you care about here is \(f(x)=x,\) and the point you care about is \(x=1\). You already know that \(f(1)=1\), so your proposed limit is \(L=1\). You are given the value for \(\epsilon\). If you draw in pink dotted lines representing \(y=L+\epsilon\) and \(y=L-\epsilon\), that gives you the \(\epsilon\) window around the proposed limit \(L\).
Fig. 1. Proposed limit for \(f(x)=x\) .
By picking a small enough \(\delta\), you can draw in the dotted lines representing \(x=1-\delta\) and \(x=1+\delta\). As long as the \(x\) value you are looking at is between those two vertical dotted lines (the \(\delta\) window), you can guarantee that the function values are within the \(\epsilon\) window. That means your choice of the limit was the correct one.
Does it matter if there is a single point that is different?
Changing the previous example just a little, define the function by
\[f(x) =\left\lbrace \begin{align} x \quad x\neq 1 \\ 0.6 \quad x=1 \end{align}\right. \]
So what should your proposed limit be? Should it be \(L=1\) or \(L=0.6\)?
The question to ask yourself is if the \(x\) values are close to \(a=1\), what are the function values close to? In this case, the function values are close to \(L=1\), so that is the proposed limit. As you can see with the exception of that one point on the graph \((1,0.6)\), all of the function values are certainly within the \(\epsilon\) window as long as the \(x\) values are within the \(\delta\) window.
Fig. 2. Limit of a function with a displaced point.
So it seems like having the function value be different at only one point shouldn't make a difference to the value of the limit.
Does it matter if the function is defined at the point?
Let's change the previous example slightly. In this case the function is \(f(x)=x\) if \(x\neq 1\), and the function isn't even defined for \(x=1\). Just like in the previous two examples, as long as the \(x\) values are close to \(a=1\), the function values are close to the proposed limit \(L=1\), so it seems that the limit of the function should still be \(1\).
Fig. 3. Limit of a function with a hole in it.
Definition of the Limit of a Function
So when you decide on the definition of the limit, from the previous examples you can see that the function value at \(x=a\) shouldn't matter. Indeed, using \(0<|x-a|<\delta\) excludes the point \(x=a\) so the limit is properly defined even if the function has a hole in it.
Let \(f\) be a function defined on the set \(S\) except possibly the point \(a\) in the interior of \(S\). We say that the limit as \(x\) approaches \(a\) of \(f(x)\) equals the number \(L\) if for all \(\epsilon >0\), there exists a \(\delta >0\) such that \(0<|x-a|<\delta\) implies \(|f(x)-L|<\epsilon\).
The notation for the limit of a function is generally
\[\lim\limits_{x\to a}f(x)=L.\]
This is read as "the limit of the function as \(x\) approaches \(a\) is equal to \(L\)".
In the case where you can't find a \(\delta\) that works at the point \(a\) you say the limit doesn't exist there.
Next let's look at the steps of applying the definition of the limit for a specific function using a graph and table.
Graphing to Find the Limit of a Function
For this example take the function to be
\[f(x)=\frac{1}{4}(x+1)(x-1)(x-5),\]
and find the limit as \(x\to 3\).
Step 1. Make a proposal for what the limit of the function should be. To do this, graph the function and make a table of the function values near \(x=3\).
| \(x\) | \(f(x)\) |
| \(2.5\) | \(-3.28\) |
| \(2.55\) | \(-3.37\) |
| \(2.6\) | \(-3.46\) |
| \(2.65\) | \(-3.54\) |
| \(2.7\) | \(-3.62\) |
| \(2.75\) | \(-3.69\) |
| \(2.8\) | \(-3.76\) |
| \(2.85\) | \(-3.83\) |
| \(2.9\) | \(-3.89\) |
| \(2.95\) | \(-3.95\) |
| \(3.0\) | \(-4.0\) |
| \(3.05\) | \(-4.05\) |
| \(3.1\) | \(-4.09\) |
| \(3.15\) | \(-4.13\) |
| \(3.2\) | \(-4.16\) |
| \(3.25\) | \(-4.18\) |
| \(3.3\) | \(-4.22\) |
| \(3.35\) | \(-4.22\) |
| \(3.45\) | \(-3.46\) |
Table 1. Data of the function \(\frac{1}{4}(x+1)(x-1)(x-5)\) around the point \(x=3\).
Fig. 4. Graph and Table of a Function near \(x=3\).
Make a proposal for the limit as \(x\rightarrow 3\). By looking at the table and graph above, you can see that the function value at \(x=3\) is \(-4\), and that as long as the \(x\) values are close to \(3\), the function values seem to be close to \(-4\). So \(L=-4\) is a good candidate for the limit.
Step 2. Someone gives you a value of \(\epsilon\), and you need to find the corresponding \(\delta\) value. To do that, you will need to find the range of function values that are close to your proposed limit \(L\). Since you only care about \(x\) close to \(3\) and function values close to \(L\), draw in the dotted lines corresponding to \(y=L+\epsilon\) and \(y=L-\epsilon\).
| \(x\) | \(f(x)\) |
| \(2.5\) | \(-3.38\) |
| \(2.55\) | \(-3.37\) |
| \(2.6\) | \(-3.46\) |
| \(2.65\) | \(-3.54\) |
| \(2.7\) | \(-3.62\) |
| \(2.75\) | \(-3.69\) |
| \(2.8\) | \(-3.76\) |
| \(2.85\) | \(-3.83\) |
| \(2.9\) | \(-3.89\) |
| \(2.95\) | \(-3.95\) |
| \(3.0\) | \(-4\) |
| \(3.05\) | \(-4.05\) |
| \(3.1\) | \(-4.09\) |
| \(3.15\) | \(-4.13\) |
| \(3.2\) | \(-4.16\) |
| \(3.25\) | \(-4.18\) |
| \(3.3\) | \(-4.2\) |
| \(3.35\) | \(-4.22\) |
| \(3.4\) | \(-4.22\) |
| \(3.45\) | \(-3.46\) |
Table 2. Data of the function \(\frac{1}{4}(x+1)(x-1)(x-5)\) around the point \(x=3\).
Fig. 5. Graph and Table of a Function near \(x=3\), larger size.
Step 3. Now you can see that you really want to zoom in on the graph to get a better idea of what is happening. It can also help to add a column to your table showing how close you are to the proposed limit.
| \(x\) | \(f(x)\) | \(|f(x)-L|\) |
| \(2.8\) | \(-3.76\) | \(0.24\) |
| \(2.85\) | \(-3.38\) | \(0.17\) |
| \(2.9\) | \(-3.89\) | \(-0.11\) |
| \(2.95\) | \(-3.95\) | \(0.05\) |
| \(3\) | \(-4\) | \(0\) |
| \(3.05\) | \(-4.05\) | \(0.05\) |
| \(3.1\) | \(-4.09\) | \(0.09\) |
| \(3.15\) | \(-4.13\) | \(0.13\) |
| \(3.2\) | \(-4.16\) | \(0.16\) |
Table 3. Data of the function \(\frac{1}{4}(x+1)(x-1)(x-5)\) arund the point \(x=3\) in a smaller interval.
Fig. 6. Zoom in on limit graph and table of function values.
Looking at the table of values, you can see that \(|f(x)-L|\) is smaller when you are to the right of \(a=3\) than if you are to the left of \(a=3\). So you can think of having two different values of "how close" you need to be to \(a=3\) to be in the \(\epsilon\) window. Call those two different values \(\delta_1\) and \(\delta_2\), and graph those lines as well.
Fig. 6. Zoom in on limit graph and table of function values.
Fig. 7. Graph of function showing epsilon and delta windows.
Step 4. You have two different candidates for what \(\delta\) should be, what do you do? Since you want both
\[0<|x-3|<\delta_1\quad \text {and} \quad 0<|x-3|<\delta_2.\]
For this to be true, take \(\delta\) to be smaller than both \(\delta_1\) and \(\delta_2\). This will be the \(\delta\) you need to finish proving that
\[\lim\limits_{x\rightarrow 3} f(x)=-4.\]
So the steps to proving that you have found the limit of a function using the definition are,
Step 1. Make a proposal for what you think the limit should be, and call it \(L\).
Step 2. Using the value of \(\epsilon\) that is given to you, find the range of the function \(f(x)\) between the lines \(y=L+\epsilon\) and \(y=L-\epsilon\).
Step 3. Find values of \(\delta\), or possibly two different values depending on the function, that guarantee that if you are within those \(\delta\) values of \(x=a\) then the function values are within the \(\epsilon\) window of \(L\). Do this by looking at \(|f(x)-L|\).
Step 4. Take the smallest of the two values for \(\delta\) you found, and write the proof up in a pretty way.
Examples of the Limit of a Function
Let's look at some examples of applying the definition to find the limit of a function.
Take \(f(x)=k\) where \(a\) and \(k\) are constant real numbers. Show that\[\lim\limits_{x\rightarrow a} f(x)=k.\]
Solution
You are given the candidate for the limit in this example. Then using the definition, for any \(\epsilon>0\) you are given,
\[|f(x)-k|=|k-k|=0<\epsilon.\]
In fact, it doesn't matter what \(\delta\) you pick, so take \(\delta>0\) to be some fixed number. Then
\[0<|x-a|<\delta\]
definitely implies that
\[|f(x)-k|=|k-k|=0<\epsilon,\]
so
\[\lim\limits_{x\to a} f(x)=\lim\limits_{x\to a} k=k.\]
Generally you won't be able to take \(\delta\) to be anything you like! It usually depends on the function, the value of \(a\), and the \(\epsilon\) in question.
Take \(f(x)=2x-3\), and let \(a=7\). Show that \[\lim\limits_{x\to 7}f(x)=11.\]
Solution
Step 1.
This is already done since you have the candidate for the limit given to you.
Step 2.
You are given an \(\epsilon>0\). The two lines you want the function values to be between are
\[y=L-\epsilon=11-\epsilon\quad \text{and}\quad y=11+\epsilon.\]
Step 3.
Look at \(|f(x)-L|\). So
\[|f(x)-L|=|2x-3-11|=|2x-14|\]
You want to find a \(\delta>0\) so that if you know that
\[0<|x-7|<\delta, \quad \text{then} \quad |2x-14|<\epsilon.\]
Let's do a little factoring,
\[|2x-14|=2|x-7|.\]
So you definitely want \(2|x-7|<\epsilon\), or if you divide by 2, you want
\[|x-7|<\frac{\epsilon}{2}.\]
That is really handy! It means if you take
\[\delta <\frac{\epsilon}{2},\]
then you know that,
\[0<|x-7|<\delta\]
implies
\[|2x-14|=2|x-7|<2\delta<2\frac{\epsilon}{2}=\epsilon\]
Step 4.
Now all that is left is to write your proof in a pretty way. It will look something like this,
Given \(\epsilon>0,\) take \[0<\delta<\frac{\epsilon}{2}.\]
Then \(0<|x-7|<\delta\) implies
\[|2x-14|=2|x-7|<2\delta<2\frac{\epsilon}{2}=\epsilon.\]
Therefore
\[\lim\limits_{x\to 7} f(x)=11. \]
So as you can see, the value of \(\delta\) in the previous example did depend on \(a=7, f(7)\) and \(\epsilon.\)
Limits of a Function - Key takeaways
- Let \(f\) be a function defined on the set \(S\) except possibly the point \(a\) in the interior of \(S\) . We say that the limit as \(x\) approaches \(a\) of \(f(x)\) equals the number \(L\) if for all \(\epsilon>0\), there exists a \(\delta>0\) such that \(0<|x-a|<\delta\) implies \(|f(x)-L|<\epsilon.\)
- The notation for the limit of a function is generally
\[\lim\limits_{x\to a} f(x)=L.\]
This is read as "the limit of the function as \(x\) approaches \(a\) is equal to \(L\)".
Finding a candidate for the limit of a function can be done using a graph or table.
The steps to proving that you have found the limit of a function using the definition are,
Step 1. Make a proposal for what you think the limit should be, and call it \(L\).
Step 2. Using the value of \(\epsilon\) that is given to you, find the range of the function between the lines \(y=L+\epsilon\) and \(y=L-\epsilon.\)
Step 3. Find values of \(\delta\), or possibly two different values depending on the function, that guarantee that if you are within those \(\delta\) values of \(x=a\) then the function values are within the \(\epsilon\) window of \(L\). Do this by looking at \(|f(x)-L|.\)
Step 4.Take the smallest of the two values for \(\delta\) you found, and write the proof up in a pretty way.
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