Function Limits III: The Gory Details
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The goal in this lesson is to provide a mathematical definition of a function limit. The idea here is that as we get closer and closer to $a$ the function gets closer and closer to $L$. To do this we will choose how close we want the function to get to $L$ by specifying an amount we call $\epsilon$. We define a variable $\delta$ to tell us how close to $a$ we need to be to accomplish this. Here is an important picture to keep in your head:
Now for the rigorous mathematical definition:
Definition of a Function Limit |
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When we say \begin{equation} \lim_{x\to a}\,f(x)=L \end{equation} we mean the following: (There is one subtlety we mention here that you should absolutely not concern yourself with if you are new to function limits. We mention it only for math majors.) |
Wow! Say that ten times fast. That is a mouthful so lets break it down into some simple steps to show what this means. In its most simplest form, we are wanting to find a relationship between $\delta$ and $\epsilon$. We can think of $\delta$ as a function of $\epsilon$ and try to find $\delta(\epsilon)$. So in short, we are going to try and make a guess about what $\delta(\epsilon)$ is and then see if it satisfies our definition of a function limit. To make this educated guess we will try to find a relationship between $|x-a|$ and $|f(x)-L|$.
Let's do a specific example to get started. Let's prove that
\begin{equation}
\lim_{x\to 0}\,m\,x+b=b.
\end{equation}
You should recognize the function as the straight line and that we are trying to show that the straight line's function limit as it approaches zero is its y-intercept. This seems pretty obvious to us because we just plug zero into the straight line and get $b$, but to actually prove it we need to follow our definition. To make an educated guess about $\delta(\epsilon)$ we want to see if there is a relationship between $|x-a|$ and $|f(x)-L|$. In this particular example $|x-a|=|x|$ and $|f(x)-L|=|m\,x+b-b|$. We can see very quickly that
\begin{eqnarray}
|f(x)-L|&=&|m\,x+b-b|\\
&=&|m\,x|\\
&=&|m|\,|x|.\\
\end{eqnarray}
So now that we know that $|m\,x+b-b|=|m|\,|x|$ how do we make our educated guess for $\delta(\epsilon)$? Let's try reconstructing what we actually want to prove
\begin{eqnarray}
|x|&\lt&\delta \\
|m|\,|x|&\lt&|m|\,\delta \\
|m\,x+b-b|&=&|m|\,|x|\lt|m|\,\delta.
\end{eqnarray}
But we remember that we also want
\begin{eqnarray}
|m\,x+b-b|&\lt&\epsilon\\
|m\,x+b-b|&=&|m|\,|x|\lt\epsilon. \\
\end{eqnarray}
Since $|m|\,|x|\lt|m|\,\delta$ and $|m|\,|x|\lt\epsilon$ why don't we try $\delta(\epsilon)=\epsilon/|m|$? Note that it may appear that we have proven something, but we haven't. All we have done is to use some math to make an educated guess. We haven't proven anything yet!
Now that we have our guess of $\delta(\epsilon)=\epsilon/|m|$ we want to see if it satisfies the definition. Remember that we are wanting to show that for every $\epsilon$ there is a $\delta$ such that if $0\lt |x-a|\lt\delta$, then $|f(x)-L|\lt\epsilon$. Here is how to do it:
- Pick some positive number for $\epsilon$. Say $\epsilon=C$.
- Find the value of $\delta$ from $\delta(\epsilon)$. In this case we have $\delta=C/|m|$
- Now assume that we are at some value for $x$ such that $0\lt |x-a|\lt\delta$. In this case we have $0\lt |x|\lt\delta$.
- Now plug the function $\delta(\epsilon)$ into $|x-a|\lt\delta$. For this problem we have $|x|\lt C/|m|$.
- Now do some basic algebra to get $|x-a|\lt\delta(\epsilon)$ into the form $|f(x)-L|\lt\epsilon$. Again, for this particular example we have
\begin{eqnarray}
|x|&\lt& C/|m|\\
|m|\,|x|&\lt&C\\
|m\,x|&\lt&C\\
|m\,x+b-b|&\lt&C\\
\end{eqnarray} - Because $C$ is an arbitrary variable, the algebra above holds true for any value of $C$ and we have therefore proven that for any value of $\epsilon$ there is a $\delta$ ($\delta(\epsilon)=\epsilon/|m|$) such that when $0\lt |x|\lt\delta$ then $|m\,x+b-b|\lt\epsilon$.
But what if we had made a bad guess? How would that have looked? Let's say we guessed $\delta(\epsilon)=10$. Let's go through the steps again:
- Pick some positive number for $\epsilon$. Say $\epsilon=C$.
- Find the value of $\delta$ from $\delta(\epsilon)$. In this case we have $\delta=10$
- Now assume that we are at some value for $x$ such that $|x-a|\lt\delta$. In this case we have $|x|\lt\delta$.
- Now plug the function $\delta(\epsilon)$ into $0\lt |x-a|\lt\delta$. For this problem we have $0\lt |x|\lt 10$.
- Now do some basic algebra to get $|x-a|\lt\delta(\epsilon)$ into the form $|f(x)-L|\lt\epsilon$. Again, for this particular example we have
\begin{eqnarray}
|x|&\lt& 10\\
|m|\,|x|&\lt&10\,|m|\\
|m\,x|&\lt&10\,|m|\\
|m\,x+b-b|&\lt&10\,|m|\\
\end{eqnarray} - Because $C$ is an arbitrary variable, I can find a value of $C$ where $C\lt 10\,|m|$. When I do this I am not guaranteed that $|m\,x+b-b|\lt\epsilon=C$. The algebra above show that when $|x|\lt 10$ then $|m\,x+b-b|\lt10\,|m|$. But from the definition when $|x|\lt 10$ then $|m\,x+b-b|$ should be less then $\epsilon$ but it isn't. See the figure below.
In the first lesson we talked about how the function could have a hole in it and
\begin{equation}
\lim_{x\to a}\,f(x)=L
\end{equation}
still be true. Our rigorous definition takes this into account in a very simple way. The idea that the function can have a hole in it is in the part of the definition where we say that $0\lt |x-a|\lt\delta$. In particular, it is the $0\lt |x-a|$ part that allows for the discontinuities by not requiring $x=a$ and thus not requiring that we know what the function is at $x=a$.
Here is a list summarizing our steps:
Step | Description |
---|---|
1 | Make an educated guess for $\delta(\epsilon)$ by finding a relationship between $|x-a|$ and $|f(x)-L|$. |
2 | Pick some positive number for $\epsilon$. Say $\epsilon=C$. |
3 | Find the value of $\delta$ from $\delta(\epsilon)$. |
4 | Now assume that we are at some value for $x$ such that $0\lt |x-a|\lt\delta$. |
5 | Now plug the function $\delta(\epsilon)$ into $|x-a|\lt\delta$. |
6 | Now do some basic algebra to get $|x-a|\lt\delta(\epsilon)$ into the form $|f(x)-L|\lt D$. The value of $D$ will naturally appear when you put $|x-a|\lt\delta(\epsilon)$ in the form $|f(x)-L|\lt D$. |
7 | Remember that $C$ can take on any value. Now do a comparison between $C$ and $D$. If $D$ is equal to or smaller than $C$ then you have made a correct guess for $\delta(\epsilon)$ and the proof is complete. If $D$ is larger than $C$ then the proof does not hold and you did not make a correct guess for $\delta(\epsilon)$. If it is impossible to find a $\delta$ for even one $\epsilon$ then you have to conclude that \begin{equation} \lim_{x\to a}\,f(x)\neq L. \end{equation} This is because our definition requires us to find a $\delta$ for every value of $\epsilon$. In this situation, the function either has a limit as $x\to a$ and it is a value different from $L$ or the function limit does not exist at all. |
It is important to note that the above steps don't tell us what the value of $L$ is. Once we make a guess for what the number $L$ is in
\begin{equation}
\lim_{x\to a}\,f(x)=L
\end{equation}
then the steps allow us to see if it is true or not. Determining a function limit can be obvious and intuitive for many functions, but it still needs to be proven with the steps above. However, there are many functions for which function limits are not intuitive. This section does not address in any way how to determine function limits. It only addresses how to determine if a guess about a function limit is true or not. Typically, you are not required to determine a function limit in most beginning courses on Calculus outside of plugging in $a$ to find $f(a)$. You will be given a function limit and asked to determine if it is true or not.
See Function Limits VII: Putting It All Together With Useful Examples.