2.2: Limits of Functions
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 10099
The concept of a limit or limiting process, essential to the understanding of calculus, has been around for thousands of years. In fact, early mathematicians used a limiting process to obtain better and better approximations of areas of circles. Yet, the formal definition of a limit—as we know and understand it today—did not appear until the late 19th century. We therefore begin our quest to understand limits, as our mathematical ancestors did, by using an intuitive approach. At the end of this chapter, armed with a conceptual understanding of limits, we examine the formal definition of a limit.
We begin our exploration of limits by taking a look at the graphs of the functions
 \(f(x)=\frac{x^2−4}{x−2}\),
 \(g(x)=\dfrac{x−2}{x−2}\), and
 \(h(x)=\frac{1}{(x−2)^2}\),
which are shown in Figure \(\PageIndex{1}\). In particular, let’s focus our attention on the behavior of each graph at and around \(x=2\).
Figure \(\PageIndex{1}\): These graphs show the behavior of three different functions around \(x=2\).
Each of the three functions is undefined at \(x=2\), but if we make this statement and no other, we give a very incomplete picture of how each function behaves in the vicinity of \(x=2\). To express the behavior of each graph in the vicinity of 2 more completely, we need to introduce the concept of a limit.
Intuitive Definition of a Limit
Let’s first take a closer look at how the function \(f(x)=(x^2−4)/(x−2)\) behaves around \(x=2\) in Figure \(\PageIndex{1}\). As the values of x approach 2 from either side of 2, the values of \(y=f(x)\) approach 4. Mathematically, we say that the limit of \(f(x)\) as x approaches 2 is 4. Symbolically, we express this limit as
\(\displaystyle \lim_{x \to 2} f(x)=4\).
From this very brief informal look at one limit, let’s start to develop an intuitive definition of the limit. We can think of the limit of a function at a number a as being the one real number \(L\) that the functional values approach as the xvalues approach a, provided such a real number L exists. Stated more carefully, we have the following definition:
Definition (Intuitive): Limit
Let \(f(x)\) be a function defined at all values in an open interval containing \(a\), with the possible exception of a itself, and let \(L\) be a real number. If all values of the function \(f(x)\) approach the real number \(L\) as the values of \(x(≠a)\) approach the number a, then we say that the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\). (More succinct, as \(x\) gets closer to \(a\), \(f(x)\) gets closer and stays close to \(L\).) Symbolically, we express this idea as
\[\lim_{x \to a} f(x)=L.\]
We can estimate limits by constructing tables of functional values and by looking at their graphs. This process is described in the following ProblemSolving Strategy.
ProblemSolving Strategy: Evaluating a Limit Using a Table of Functional Values
1. To evaluate \(\displaystyle \lim_{x \to a} f(x)\), we begin by completing a table of functional values. We should choose two sets of xvalues—one set of values approaching a and less than a, and another set of values approaching a and greater than \(a\). Table \(\PageIndex{1}\) demonstrates what your tables might look like.
\(x\)  \(f(x)\)  \(x\)  \(f(x)\) 

\(a0.1\)  \(f(a0.1)\)  \(a+0.1\)  \(f(a+0.1)\) 
\(a0.01\)  \(f(a0.01)\)  \(a+0.001\)  \(f(a+0.001)\) 
\(a0.001\)  \(f(a0.001)\)  \(a+0.0001\)  \(f(a+0.001)\) 
\(a0.0001\)  \(f(a0.0001)\)  \(a+0.00001\)  \(f(a+0.0001)\) 
Use additional values as necessary.  Use additional values as necessary. 
2. Next, let’s look at the values in each of the \(f(x)\) columns and determine whether the values seem to be approaching a single value as we move down each column. In our columns, we look at the sequence \(f(a−0.1)\), \(f(a−0.01)\), \(f(a−0.001)\), \(f(a−0.0001)\), and so on, and \(f(a+0.1),f(a+0.01),f(a+0.001),f(a+0.0001)\), and so on. (Note: Although we have chosen the xvalues \(a±0.1,a±0.01,a±0.001,a±0.0001\), and so forth, and these values will probably work nearly every time, on very rare occasions we may need to modify our choices.)
3. If both columns approach a common yvalue L, we state \(\displaystyle \lim_{x \to a}f(x)=L\). We can use the following strategy to confirm the result obtained from the table or as an alternative method for estimating a limit.
4. Using a graphing calculator or computer software that allows us graph functions, we can plot the function \(f(x)\), making sure the functional values of \(f(x)\) for xvalues near a are in our window. We can use the trace feature to move along the graph of the function and watch the yvalue readout as the xvalues approach a. If the yvalues approach L as our xvalues approach a from both directions, then \(\displaystyle \lim_{x \to a}f(x)=L\). We may need to zoom in on our graph and repeat this process several times.
We apply this ProblemSolving Strategy to compute a limit in Examples \(\PageIndex{1A}\) and \(\PageIndex{1B}\).
Example \(\PageIndex{1A}\): Evaluating a Limit Using a Table of Functional Values
Evaluate \(\displaystyle \lim_{x \to 0}\frac{\sin x}{x}\) using a table of functional values.
Solution
We have calculated the values of \(f(x)=(\sin x)/x\) for the values of \(x\) listed in Table \(\PageIndex{2}\).
\(x\)  \(\frac{\sin x}{x}\)  \(x\)  \(\frac{\sin x}{x}\) 

0.1  0.998334166468  0.1  0.998334166468 
0.01  0.999983333417  0.01  0.999983333417 
0.001  0.999999833333  0.001  0.999999833333 
0.0001  0.999999998333  0.0001  0.999999998333 
Note: The values in this table were obtained using a calculator and using all the places given in the calculator output.
As we read down each \(\frac{(\sin x)}{x}\) column, we see that the values in each column appear to be approaching one. Thus, it is fairly reasonable to conclude that \(\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1\). A calculatoror computergenerated graph of \(f(x)=\frac{(sinx)}{x}\) would be similar to that shown in Figure \(\PageIndex{2}\), and it confirms our estimate.
Figure \(\PageIndex{2}\): The graph of \(f(x)=(\sin x)/x\) confirms the estimate from Table.
Example \(\PageIndex{1B}\): Evaluating a Limit Using a Table of Functional Values
Evaluate \(\displaystyle \lim_{x\to4}\frac{\sqrt{x}−2}{x−4}\) using a table of functional values.
Solution
As before, we use a table—in this case, Table \(\PageIndex{3}\)—to list the values of the function for the given values of \(x\).
\(x\)  \(\frac{\sqrt{x}−2}{x−4}\)  \(x\)  \(\frac{\sqrt{x}−2}{x−4}\) 

3.9  0.251582341869  4.1  0.248456731317 
3.99  0.25015644562  4.01  0.24984394501 
3.999  0.250015627  4.001  0.249984377 
3.9999  0.250001563  4.0001  0.249998438 
3.99999  0.25000016  4.00001  0.24999984 
After inspecting this table, we see that the functional values less than 4 appear to be decreasing toward 0.25 whereas the functional values greater than 4 appear to be increasing toward 0.25. We conclude that \(\displaystyle \lim_{x\to4}\frac{\sqrt{x}−2}{x−4}=0.25\). We confirm this estimate using the graph of \(f(x)=\frac{\sqrt{x}−2}{x−4}\) shown in Figure \(\PageIndex{3}\).
Figure \(\PageIndex{3}\): The graph of \(\frac{\sqrt{x}−2}{x−4}\) confirms the estimate from table
Exercise \(\PageIndex{1}\)
Estimate \(\displaystyle \lim_{x \to 1} \frac{\frac{1}{x}−1}{x−1}\) using a table of functional values. Use a graph to confirm your estimate.
 Hint

Use 0.9, 0.99, 0.999, 0.9999, 0.99999 and 1.1, 1.01, 1.001, 1.0001, 1.00001 as your table values.
 Answer

\[\lim_{x\to1}\frac{\frac{1}{x}−1}{x−1}=−1\]
At this point, we see from Examples \(\PageIndex{1A}\) and \(\PageIndex{1b}\) that it may be just as easy, if not easier, to estimate a limit of a function by inspecting its graph as it is to estimate the limit by using a table of functional values. In Example \(\PageIndex{2}\), we evaluate a limit exclusively by looking at a graph rather than by using a table of functional values.
Example \(\PageIndex{2}\): Evaluating a Limit Using a Graph
For \(g(x)\) shown in Figure \(\PageIndex{4}\), evaluate \(\displaystyle \lim_{x\to−1}g(x)\).
Figure \(\PageIndex{4}\): The graph of \(g(x)\) includes one value not on a smooth curve.
Solution:
Despite the fact that \(g(−1)=4\), as the xvalues approach −1 from either side, the \(g(x)\) values approach 3. Therefore, \(\displaystyle \lim_{x\to−1}g(x)=3\). Note that we can determine this limit without even knowing the algebraic expression of the function.
Based on Example \(\PageIndex{2A}\), we make the following observation: It is possible for the limit of a function to exist at a point, and for the function to be defined at this point, but the limit of the function and the value of the function at the point may be different.
Exercise \(\PageIndex{2}\)
Use the graph of \(h(x)\) in Figure \(\PageIndex{5}\) to evaluate \(\displaystyle \lim_{x \to 2}h(x)\), if possible.
Figure \(\PageIndex{5}\):
 Hint

What yvalue does the function approach as the xvalues approach 2?
 Solution

\(\displaystyle \lim_{x \to 2}h(x)=−1.\)
Looking at a table of functional values or looking at the graph of a function provides us with useful insight into the value of the limit of a function at a given point. However, these techniques rely too much on guesswork. We eventually need to develop alternative methods of evaluating limits. These new methods are more algebraic in nature and we explore them in the next section; however, at this point we introduce two special limits that are foundational to the techniques to come.
Two Important Limits
Let a be a real number and c be a constant.
 \(\displaystyle \lim_{x \to a}x=a\)
 \(\displaystyle \lim_{x \to a}c=c\)
We can make the following observations about these two limits.
 For the first limit, observe that as x approaches a, so does \(f(x)\), because \(f(x)=x\). Consequently, \(\displaystyle \lim_{x \to a}x=a\).
 For the second limit, consider Table.
\(x\)  \(f(x0=c\)  \(x\)  \(f(x)=c\) 

\(a0.1\)  c  \(a+0.1\)  c 
\(a0.01\)  c  \(a+0.01\)  c 
\(a0.001\)  c  \(a+0.001\)  c 
\(a0.0001\)  c  \(a+0.0001\)  c 
Observe that for all values of x (regardless of whether they are approaching a), the values \(f(x)\) remain constant at c. We have no choice but to conclude \(\displaystyle \lim_{x \to a}c=c\).
The Existence of a Limit
As we consider the limit in the next example, keep in mind that for the limit of a function to exist at a point, the functional values must approach a single realnumber value at that point. If the functional values do not approach a single value, then the limit does not exist.
Example \(\PageIndex{3}\): Evaluating a Limit That Fails to Exist
Evaluate \(\displaystyle\lim_{x \to 0}\sin(1/x)\) using a table of values.
Solution
Table \(\PageIndex{3}\) lists values for the function \(\sin(1/x)\) for the given values of \(x\).
\(x\)  \(\sin(1/x)\)  \(x\)  \(\sin(1/x)\) 

0.1  0.544021110889  0.1  −0.544021110889 
0.01  0.50636564111  0.01  −0.50636564111 
0.001  −0.8268795405312  0.001  0.8268795405312 
0.0001  0.305614388888  0.0001  −0.305614388888 
0.00001  −0.035748797987  0.00001  0.035748797987 
0.000001  0.349993504187  0.000001  −0.349993504187 
After examining the table of functional values, we can see that the yvalues do not seem to approach any one single value. It appears the limit does not exist. Before drawing this conclusion, let’s take a more systematic approach. Take the following sequence of xvalues approaching 0:
\[\frac{2}{π},\frac{2}{3π},\frac{2}{5π},\frac{2}{7π},\frac{2}{9π},\frac{2}{11π},….\]
The corresponding yvalues are
\[1,1,1,1,1,1,....\]
At this point we can indeed conclude that \(\displaystyle \lim_{x \to 0}sin(1/x)\) does not exist. (Mathematicians frequently abbreviate “does not exist” as DNE. Thus, we would write \(\displaystyle \lim_{x \to 0}\sin(1/x)\) DNE.) The graph of \(f(x)=sin(1/x)\) is shown in Figure \(\PageIndex{6}\) and it gives a clearer picture of the behavior of \(sin(1/x)\) as x approaches 0. You can see that \(sin(1/x)\) oscillates ever more wildly between −1 and 1 as \(x\) approaches 0.
Figure \(\PageIndex{6}\): The graph of \(f(x)=\sin (1/x)\) oscillates rapidly between −1 and 1 as \(x\) approaches 0.
Exercise \(\PageIndex{3}\)
Use a table of functional values to evaluate \(\displaystyle \lim_{x \to 2}\frac{∣x^2−4∣}{x−2}\), if possible.
 Hint

Use xvalues 1.9, 1.99, 1.999, 1.9999, 1.9999 and 2.1, 2.01, 2.001, 2.0001, 2.00001 in your table.
 Answer

\(\displaystyle \lim_{x \to 2}\frac{∣x^2−4∣}{x−2}\) does not exist.
OneSided Limits
Sometimes indicating that the limit of a function fails to exist at a point does not provide us with enough information about the behavior of the function at that particular point. To see this, we now revisit the function \(g(x)=x−2/(x−2)\) introduced at the beginning of the section (see Figure(b)). As we pick values of x close to 2, \(g(x)\) does not approach a single value, so the limit as x approaches 2 does not exist—that is, \(\displaystyle \lim_{x \to 2}g(x)\) DNE. However, this statement alone does not give us a complete picture of the behavior of the function around the xvalue 2. To provide a more accurate description, we introduce the idea of a onesided limit. For all values to the left of 2 (or the negative side of 2), \(g(x)=−1\). Thus, as x approaches 2 from the left, \(g(x)\) approaches −1. Mathematically, we say that the limit as x approaches 2 from the left is −1. Symbolically, we express this idea as
\[\lim_{x \to 2^−}g(x)=−1.\]
Similarly, as x approaches 2 from the right (or from the positive side), \(g(x)\) approaches 1. Symbolically, we express this idea as
\[\lim_{x \to 2^+}g(x)=1.\]
We can now present an informal definition of onesided limits.
Definition: Onesided Limits
We define two types of onesided limits.
Limit from the left:
Let \(f(x)\) be a function defined at all values in an open interval of the form \(z\), and let \(L\) be a real number. If the values of the function \(f(x)\) approach the real number \(L\) as the values of \(x\) (where \(x<a\)) approach the number a, then we say that \(L\) is the limit of \(f(x)\) as \(x\) approaches a from the left. Symbolically, we express this idea as
\[\lim_{x \to a^−}f(x)=L.\]
Limit from the right:
Let \(f(x)\) be a function defined at all values in an open interval of the form \((a,c)\), and let \(L\) be a real number. If the values of the function \(f(x)\) approach the real number \(L\) as the values of \(x\) (where \(x>a\)) approach the number \(a\), then we say that \(L\) is the limit of \(f(x)\) as x approaches a from the right. Symbolically, we express this idea as
\[\lim_{x \to a^+}f(x)=L.\]
Example \(\PageIndex{4}\): Evaluating OneSided Limits
For the function \(f(x)=\begin{cases}x+1& \text{if }x<2\\ x^2−4 & \text{if }x≥2\end{cases}\), evaluate each of the following limits.
 \(\displaystyle \lim_{x \to 2^−}f(x)\)
 \(\displaystyle \lim_{x \to 2^+}f(x)\)
Solution
We can use tables of functional values again Table. Observe that for values of x less than 2, we use \(f(x)=x+1\) and for values of x greater than 2, we use \(f(x)=x^2−4.\)
\(x\)  \(f(x)=x+1\)  \(x\)  \(f(x)=x^24\) 

1.9  2.9  2.1  0.41 
1.99  2.99  2.01  0.0401 
1.999  2.999  2.001  0.004001 
1.9999  2.9999  2.0001  0.00040001 
1.99999  2.99999  2.00001  0.0000400001 
Based on this table, we can conclude that a. \(\displaystyle \lim_{x \to 2^−}f(x)=3\) and b. \(\displaystyle \lim_{x \to 2^+}f(x)=0\). Therefore, the (twosided) limit of \(f(x)\) does not exist at \(x=2\). Figure \(\PageIndex{7}\) shows a graph of \(f(x)\) and reinforces our conclusion about these limits.
Figure \(\PageIndex{7}\): The graph of \(f(x)=\begin{cases}x+1& \text{if }x<2\\ x^2−4 & \text{if }x≥2\end{cases}\) has a break at x=2.
Exercise \(\PageIndex{4}\)
Use a table of functional values to estimate the following limits, if possible.
 \(\displaystyle \lim_{x→2^−}\frac{∣x^2−4∣}{x−2}\)
 \(\displaystyle \lim_{x→2^+}\frac{∣x^2−4∣}{x−2}\)
 Hint

Use xvalues 1.9, 1.99, 1.999, 1.9999, 1.9999 to estimate \(\displaystyle \lim_{x→2^−}\frac{∣x^2−4∣}{x−2}\).
Use xvalues 2.1, 2.01, 2.001, 2.0001, 2.00001 to estimate \(\displaystyle \lim_{x→2^+}\frac{∣x^2−4∣}{x−2}.\)
(These tables are available from a previous Checkpoint problem.)
 Solution a

a. \(\displaystyle \lim_{x→2^−}\frac{∣x^2−4∣}{x−2}=−4\)
 Solution b

\(\displaystyle \lim_{x→2^+}\frac{∣x^2−4∣}{x−2}=4\)
Let us now consider the relationship between the limit of a function at a point and the limits from the right and left at that point. It seems clear that if the limit from the right and the limit from the left have a common value, then that common value is the limit of the function at that point. Similarly, if the limit from the left and the limit from the right take on different values, the limit of the function does not exist. These conclusions are summarized in Note.
Relating OneSided and TwoSided Limits
Let \(f(x)\) be a function defined at all values in an open interval containing \(a\), with the possible exception of \(a\) itself, and let L be a real number. Then,
\(\displaystyle \lim_{x \to a}f(x)=L\) if and only if \(\displaystyle \lim_{x \to a^−}f(x)=L\) and \(\displaystyle \lim_{x \to a^+} f(x)=L\).
or we could say
\(\displaystyle \lim_{x \to a}f(x)\) exists if and only if \(\displaystyle \lim_{x \to a^−}f(x)=\lim_{x \to a^+}f(x)\).
Key Equations
\(\displaystyle \lim_{x \to a}f(x)=L\)
\(\displaystyle \lim_{x \to a}x=a\) \(\displaystyle \lim_{x \to a}c=c\)
\(\displaystyle \lim_{x \to a^−}f(x)=L\) \(\displaystyle \lim_{x \to a^+}f(x)=L\)
Glossary
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CCBYSANC 4.0 license. Download for free at http://cnx.org.