Function Limits

Ok, this is one of those picky little details that you probably won't ever need to worry about unless you are a math major. So, if you are not a math major you can now go and feel comfortable that you don't need to understand anything in this lesson.

Now that you are still here there is one extremely minor detail that does not really add to our discussion of function limits, but does need to be addressed if we are to have a complete discussion of function limits. In our definition of function limits we said that
\[
\lim_{x\to a}\,f(x)=L
\]
means that for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that if $0\lt |x-a|\lt\delta$, then $|f(x)-L|\lt\epsilon$. What we did not emphasize at the time was that $x$ and $a$ are assumed to be in the domain of $f(x)$.

Now recall that intervals are sets and we have shown that when we say $0\lt |x-a|\lt\delta$ we mean the set $(a-\delta,a)\cup (a,a+\delta)$.

While $a$ is assumed to be in the domain of $f(x)$ it actually needs to be inside the domain. So what do we mean by in versus inside? Let's say that we have a function whose domain is $[1,5]$. The numbers $1$ and $5$ are in $[1,5]$. By in we mean that they are elements of the set $[1,5]$. However, $1$ and $5$ are not inside $[1,5]$. The numbers $1.238$, $3.0$ and $4.9999999$ are inside $[1,5]$, but $1$ and $5$ are not.

We require $a$ to be inside the domain so for small enough $\delta$, $(a-\delta,a)$ and $(a,a+\delta)$ can be in the domain. Let's keep with our example above and assume we have a function $f(x)$ whose domain is $[1,5]$. Now let's say we have some limit
\[
\lim_{x\to 1}\,f(x)=L
\]
that we want to prove. The problem is that we need to know that function on $(a-\delta,a)\cup (a,a+\delta)$. If $a=1$ we need to know the function on $(1-\delta,1)\cup (1,1+\delta)$. But because the domain is $[1,5]$ we can't say anything about the function in $(1-\delta,1)$!

The most that we can determine would be the limit from the right:
\[
\lim_{x\to 1^+}\,f(x)=L.
\]
Since we don't even have a number for $f(x)$ in $(1-\delta,1)$ we can't even determine a number for
\[
\lim_{x\to 1^-}\,f(x).
\]
Therefore, from this problem we know that we can't determine a value for
\[
\lim_{x\to a}\,f(x)=L.
\]

It's important to keep in mind that the definition of function limits is very highly restrictive. We need to be able to approach $a$ from both the left and the right, and as we approach $a$ from either side the function needs to approach the same value. If we can't do that then the most that we can say is that a left or right-sided limit exists.

Recall that the definition of continuity is the function limit
\[
\lim_{x\to a}\,f(x)=f(a).
\]
In our discussion of function limits we said that what we are attempting to do is find a $\delta$ that is dependent on $\epsilon$. In other words, find a $\delta(\epsilon)$ so that (now we are just repeating the definition) for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that if $0\lt |x-a|\lt\delta$, then $|f(x)-f(a)|\lt\epsilon$.

But here is the point that is very important. What we did not emphasize at the time was that $\delta$ is also dependent on $a$ as well! For example, when we proved the continuity of $x^2$, we found $\delta=\min(1,\epsilon/(2|a|+1))$. We really should have denoted it by something like $\delta_a(\epsilon)$. But now that we know what we have done, wouldn't it be nice to have a definition of continuity that was not dependent on $a$? In other words, find a $\delta$ that could work for any value of $a$. We call this Uniform Continuity and here is its definition:

Finally, there are three important things to note about uniform continuity:

• If $f(x)$ is uniformly continuous on a set $U$ and $a$ is inside $U$, then $f(x)$ is continuous at $a$.
• The straight line is a function that is continuous for every $a$ inside $U$ and also uniformly continuous on $U$. For a straight line $U$ is the entire $x$-axis. See here and here. However, this is a specific example that does not indicate how things work in general. Just because a function is continuous for every $a$ inside $U$ does not guarantee that it is uniformly continuous on $U$. This is a very subtle point best demonstrated by example. For example, we have proven here that $x^2$ is continuous on the entire $x$-axis. However, unlike a straight line, we can prove that it is not uniformly continuous on the entire $x$-axis.
• Finally, there is one situation where being continuous implies uniform continuity. If $f(x)$ is continuous on the closed interval $[a,b]$, then $f(x)$ is uniformly continuous on $[a,b]$. This result is very hard to prove but very important. The main use of this result is to prove that continuous functions are integrable.

We are assuming that you are comfortable with the gory details of proving function limits so we will just jump right into the definition:

Definitions like this can be confusing. The point is that we need to find a $Y$ for every $\epsilon$. There is no method for finding this. You can make a guess for it, but usually a mathematician will do some algebra to find a relationship between $x\gt Y$ and $|f(x)-L|\lt\epsilon$ to determine an appropriate guess. In this lesson we call this guess $Y(\epsilon)$. Once a guess is made then the definition above means the following:

The best way to get comfortable with these steps is to try several problems. The easiest one and the one you should keep in your mind is
\begin{equation}
\lim_{x\to\infty}\frac{1}{x}=0.
\end{equation}

Remember that the steps above do not tell us what $L$ is. Once we have a "guess" for $L$ then the definition of a limit proves the assertion
\begin{equation}
\lim_{x\to\infty}\,f(x)=L.
\end{equation}
In most beginning Calculus courses you usually don't have to worry about finding $L$. You will be given $f(x)$ and $L$ and you will then need to go through the steps above to prove that
\begin{equation}
\lim_{x\to\infty}\,f(x)=L.
\end{equation}

When this limit exists we say that the function has a horizontal asymptote at $L$.

We can also define $\lim_{x\to -\infty}\,f(x)=L$:

It is possible for a function to have two horizontal asymptotes: one for when $x$ goes to positive infinity and the other when $x$ goes to negative infinity. We would say that
\begin{equation}
\lim_{x\to \infty}\,f(x)=L_1
\end{equation}
and
\begin{equation}
\lim_{x\to -\infty}\,f(x)=L_2.
\end{equation}

When we are dealing with the limit of the quotient of two functions, \[\lim_{x\to a}\,\frac{f(x)}{g(x)},\] then we know that the limit does not exist when
\[
\lim_{x\to a}\,f(x)\neq 0\quad\mbox{and}\quad\lim_{x\to a}\,g(x)=0
\]

but does exist and is zero when
\[
\lim_{x\to a}\,f(x)=0\quad\mbox{and}\quad\lim_{x\to a}\,g(x)\neq 0.
\]

But what happens when both
\[
\lim_{x\to a}\,f(x)=0\quad\mbox{and}\quad\lim_{x\to a}\,g(x)=0?
\]
Can we say for certain whether the function limit exists or not? The answer is no. In this situation the function limit may exist or may not exist. The limit could be zero or unbounded. Here are important limits involving functions going to zero:

Here are some important aspects of continuous functions:

1. Continuous functions are bounded.
2. The Intermediate-Value Theorem proves our original intuition that a continuous function is one that can be drawn on a piece of paper without having to lift the pencil from the paper to draw it.
3. The Extreme-Value Theorem describes how a function will take on both a maximum value and a minimum value on $[a,b]$.
4. The composition of two continuous functions is a continuous function.

We can also use our properties of function limits to show that when $f(x)$ and $g(x)$ are continuous,
\begin{equation}
\lim_{x\to a}\,f(x)=f(a)\quad\mbox{and}\quad\lim_{x\to a}\,g(x)=g(a),
\end{equation}
then the follow properties are true:

A continuous function is one that can be drawn on a piece of paper without having to lift the pencil from the paper to draw it. For example,
\[
f(x)=\left\{\begin{array}{lr}x^2,&x\leq 1\\x,&x\gt 1\end{array}\right.
\]
is an example of a continuous function.

However,

\[
f(x)=\left\{\begin{array}{lr}x^2,&x\leq 1\\5x,&x\gt 1\end{array}\right.
\]

is a function that cannot be drawn on a piece of paper without lifting your pencil. At $x=1$ the function has a value of 1 and immediately to the right of $x=1$ the function has a value of 5! This is definitely not a continuous function because we must lift our pencil off of the sheet at $x=1$ to continue drawing the function. Note that this function is continuous at $x=2$.

Furthermore, the functions
\[
f(x)=\left\{ \begin{array}{lr}x^2, & x\lt 1\\ x^2, & x\gt 1\\ \end{array}\right.\quad\mbox{and}\quad f(x)=\left\{\begin{array}{lr}x^2,&x\lt 1\\x,&x\gt 1\end{array}\right.
\]
are both discontinuous at $x=1$ because they are not defined there at all. Note that the second function is identical to our first example except at $x=1$.

The official mathematical definition of continuity is that a function is continuous at a point $x=a$ when
\[
\lim_{x\to a}\,f(x)=f(a).
\]
We have always said that a function limit
\[
\lim_{x\to a}\,f(x)=L
\]
does not have to equal the function value at $x=a$. In other words, $L$ does not have to equal $f(a)$. But for a continuous function, $L=f(a)$. This is how we can draw a picture of a function and not have to lift up our pencil.

Similar to one-sided limits, we can define one-sided continuity:
\[
f(x)\mbox{ is continuous from the left when }\lim_{x\to a-}\,f(x)=f(a)
\]
and
\[
f(x)\mbox{ is continuous from the right when }\lim_{x\to a+}\,f(x)=f(a).
\]
But we know that $\lim_{x\to a\,}\,f(x)=L$ if and only if $\lim_{x\to a-}\,f(x)=L$ and $\lim_{x\to a+}\,f(x)=L$. Therefore, for continuous functions,
\[
\lim_{x\to a-}\,f(x)=\lim_{x\to a}\,f(x)=f(a)=\lim_{x\to a+}\,f(x).
\]
The left-sided and right-sided function limits must equal $f(a)$ for a continuous function.

Finally, here are some important properties of continuous functions.

In the lesson Function Limits II: When Function Limits Don't Exist we discussed how a function limit from the left and from the right must be equal. We introduced the notation of a function limit from left as
\begin{equation}
\lim_{x\to a\,-}\,f(x)=L_l
\end{equation}
and a function limit from the right as
\begin{equation}
\lim_{x\to a\,+}\,f(x)=L_r.
\end{equation}
We referred to these as one-sided limits. We will oftentimes refer to
\begin{equation}
\lim_{x\to a\,}\,f(x)=L
\end{equation}
as a two-sided limit.

Our goal here is to provide the "epsilon-delta" definition of these limits similar to what we did for the function limit and to show that
\begin{equation}
\lim_{x\to a\,}\,f(x)=L
\end{equation}
if and only if
\begin{equation}
\lim_{x\to a-}\,f(x)=L\quad\mbox{and}\quad\lim_{x\to a+}\,f(x)=L.
\end{equation}

Our left-hand limit is defined as

Our right-hand limit is defined as

More importantly,
\begin{equation}
\lim_{x\to a\,}\,f(x)=L
\end{equation}
if and only if
\begin{equation}
\lim_{x\to a-}\,f(x)=L\quad\mbox{and}\quad\lim_{x\to a+}\,f(x)=L.
\end{equation}
The use of "if and only if" is very important. It means two things at the same time. First, it means that if the function limit exists then we know that the left and right-sided limits exist, are equal to each other, and have the value of $L$. Second, it means that if the two sided limits exist and are equal to each other then the two-sided function limit exists and is equal to $L$. You may find the proof here.

Finally, we can prove our assertion that if the left and right-sided limits do not equal one another then the function limit does not exist.

See Function Limits VII: Putting It All Together With Useful Examples.

Here is a visual example to help you remember when function limits exists and don't exist.

We assume that
\[
\lim_{x\to a}\,f(x)=L
\]
and
\[
\lim_{x\to a}\,g(x)=K
\]
are true when discussing the properties below.