Function Limits

There are two important scenarios for which a function limit does not exist. The first is when the limit produces two different values dependent upon whether we approach a point from the left or the right. The next picture demonstrates this.

For a function limit to exist it must have the same value if we approach it from either the left or the right. Otherwise, we say that the limit does not exist.

Approaching a point from the left is referred to as a left-hand limit and has the notation
\begin{equation}
\lim_{x\to a\,-}\,f(x)=L_l.
\end{equation}
Notice the minus sign appended to $x\to a$ below $\lim$. In English we would say that "as $x$ approaches $a$ from the left, the function approaches $L_l$." Similarly, approaching a point from the right is referred to as a right-hand limit and has the notation
\begin{equation}
\lim_{x\to a\,+}\,f(x)=L_r.
\end{equation}
Notice the plus sign appended to $x\to a$ below $\lim$. In English we would say that "as $x$ approaches $a$ from the right, the function approaches $L_r$." The limit at point $a$ does not exist if the limit from the left does not equal the limit from the right. The only way the limit can exist is when $L_l=L_r$.

In the picture above, we see that
\begin{equation}
\lim_{x\to a\,-}\,f(x)=7
\end{equation}
and
\begin{equation}
\lim_{x\to a\,+}\,f(x)=4.
\end{equation}
Because the limit from the left and the limit from the right are not the same the limit at point $a$ does not exist.

The second scenario is when a function approaches infinity. See the picture below.

An example of this type of function would be
\begin{equation}
f(x)=\frac{1}{x^2-a^2}
\end{equation}
where the function gets larger and larger because the denominator gets smaller and smaller as $x\to a$. In this case we write
\begin{equation}
\lim_{x\to a}\,f(x)=\infty .
\end{equation}

We can have situations where the two scenarios above are combined to have functions like
\begin{equation}
\frac{1}{x-2}
\end{equation}
where
\begin{equation}
\lim_{x\to 2+}\,\frac{1}{x-2}=\infty
\end{equation}
and
\begin{equation}
\lim_{x\to 2-}\,\frac{1}{x-2}=-\infty.
\end{equation}
Neither the left or right sided limits are equal and even if they were the function is going to infinity! See the graph below.

Remember to keep these pictures in your head. Even though the next lesson will provide a mathematical definition of function limits, it is important to keep the pictures in your head to form some intuition about when a function limit exists.

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:

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:

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.

Function limits explain in mathematical detail how close a function gets to some number on the y-axis as it approaches a number on the x-axis. We oftentimes refer to the number on the y-axis as the function's limit and denote it by $L$. The function limit answers the question: "As $x$ approaches $a$, what does $f(x)$ intend to be?"

In the picture above, as $x$ approaches $a$, the function approaches the value $L$. In English we would say "The limit of $f(x)$ as $x$ approaches $a$ is $L$." In math we would use the notation
\begin{equation}
\lim_{x\to a}f(x)=L.
\end{equation}

It is important for you to get used to the limit notation. Differential Calculus will use function limits a lot. You could even say that Differential Calculus is the study of function limits as you will see when you begin studying derivatives.

As the function gets closer and closer to the point $a$, the function approaches a value $L$. Notice that I did not say the function approaches the value $f(a)$. The value of the function at $a$ does not have to be the same as its limit at $a$! This is an extremely important point to remember. The two pictures below demonstrate this point. In the first one we have a discontinuity at $a$ and in the second the functions does not exist at all at $a$.

So what type of function would look like this? One example would be
\begin{equation}
f(x)=\left\{
\begin{array}{r@{\quad\quad}l}
x^2, & x\lt 3 \\
1, & x=3 \\
x^2, & x>3 \\
\end{array}
\right.
.
\end{equation}
This function is just $x^2$ with a discontinuity at $x=3$. Normally the function would be 9 at $x=3$, but this function has the value of 1 at $x=3$.

Here is an example of a function with a hole in it.

If the limit at a hole and discontinuity confuses you then don't worry! I just want you to keep the pictures in your head and accept it for now. In the lesson "Function Limits III: The Gory Details" we will provide a mathematical definition of function limits and prove that the function limit can still exist even though the function has a hole or discontinuity in it. The next step is for you to understand when function limits don't exist.