2.1: Limit of a Function and Limit Laws
 Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\) Using correct notation, describe the limit of a function.
 Use a table of values to estimate the limit of a function or to identify when the limit does not exist.
 Use a graph to estimate the limit of a function or to identify when the limit does not exist.
 Recognize the basic limit laws.
 Use the limit laws to evaluate the limit of a function.
 Evaluate the limit of a function by factoring.
 Use the limit laws to evaluate the limit of a polynomial or rational function.
 Evaluate the limit of a function by factoring or by using conjugates.
 Evaluate the limit of a function by using the squeeze theorem.
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
 \(f(x)=\dfrac{x^2−4}{x−2}\),
 \(g(x)=\dfrac{x−2}{x−2}\), and
 \(h(x)=\dfrac{1}{(x−2)^2}\),
which are shown in Figure \(\PageIndex{1}\). In particular, let’s focus our attention on the behavior of each graph at and around \(x=2\).
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
\[\displaystyle \lim_{x \to 2} f(x)=4 \nonumber\]
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 \(x\)values approach \(a\), provided such a real number \(L\) exists. Stated more carefully, we have the following definition:
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
\[\lim_{x \to a} f(x)=L. \label{limit} \]
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.
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 \(x\)values—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.
\(x\)  \(f(x)\)  \(x\)  \(f(x)\) 

\(a0.1\)  \(f(a0.1)\)  \(a+0.1\)  \(f(a+0.1)\) 
\(a0.01\)  \(f(a0.01)\)  \(a+0.01\)  \(f(a+0.01)\) 
\(a0.001\)  \(f(a0.001)\)  \(a+0.001\)  \(f(a+0.001)\) 
\(a0.0001\)  \(f(a0.0001)\)  \(a+0.0001\)  \(f(a+0.0001)\) 
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 \(x\)values \(a±0.1, \;a±0.01, \;a±0.001, \;a±0.0001\), 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 \(y\)value \(L\), 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 \(x\)values near \(a\) are in our window. We can use the trace feature to move along the graph of the function and watch the \(y\)value readout as the \(x\)values approach \(a\). If the \(y\)values approach \(L\) as our \(x\)values 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 \(\PageIndex{1B}\).
Evaluate \(\displaystyle \lim_{x \to 0}\frac{\sin x}{x}\) using a table of functional values.
Solution
We have calculated the values of \(f(x)=\dfrac{\sin x}{x}\) 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 \(\dfrac{\sin x}{x}\) 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 \(f(x)=\dfrac{\sin x}{x}\) would be similar to that shown in Figure \(\PageIndex{2}\), and it confirms our estimate.
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 \(f(x)=\dfrac{\sqrt{x}−2}{x−4}\) shown in Figure \(\PageIndex{3}\).
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

\[\lim_{x\to1}\frac{\frac{1}{x}−1}{x−1}=−1\nonumber \]
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.
For \(g(x)\) shown in Figure \(\PageIndex{4}\), evaluate \(\displaystyle \lim_{x\to−1}g(x)\).
Solution:
Despite the fact that \(g(−1)=4\), as the \(x\)values 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 \(\PageIndex{2}\), 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.
Use the graph of \(h(x)\) in Figure \(\PageIndex{5}\) to evaluate \(\displaystyle \lim_{x \to 2}h(x)\), if possible.
 Hint

What \(y\)value does the function approach as the \(x\)values 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.
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 \(\PageIndex{4}\).
\(x\)  \(f(x)=c\)  \(x\)  \(f(x)=c\) 

\(a0.1\)  \(c\)  \(a+0.1\)  \(c\) 
\(a0.01\)  \(c\)  \(a+0.01\)  \(c\) 
\(a0.001\)  \(c\)  \(a+0.001\)  \(c\) 
\(a0.0001\)  \(c\)  \(a+0.0001\)  \(c\) 
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.
Evaluate \(\displaystyle\lim_{x \to 0}\sin(1/x)\) using a table of values.
Solution
Table \(\PageIndex{5}\) 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 \(y\)values 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 \(x\)values approaching \(0\):
\[\frac{2}{π},\;\frac{2}{3π},\;\frac{2}{5π},\;\frac{2}{7π},\;\frac{2}{9π},\;\frac{2}{11π},\;….\nonumber \]
The corresponding \(y\)values are
\[1,\;1,\;1,\;1,\;1,\;1,\;....\nonumber \]
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 \(0\).
Use a table of functional values to evaluate \(\displaystyle \lim_{x \to 2}\frac{∣x^2−4∣}{x−2}\), if possible.
 Hint

Use \(x\)values 1.9, 1.99, 1.999, 1.9999, 1.99999 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.
Previously, we evaluated limits by looking at graphs or by constructing a table of values. Now, we establish laws for calculating limits and learn how to apply these laws.
Evaluating Limits with the Limit Laws
The first two limit laws were stated previously and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.
For any real number \(a\) and any constant \(c\),
 \(\displaystyle \lim_{x→a}x=a\)
 \(\displaystyle \lim_{x→a}c=c\)
Evaluate each of the following limits using "Basic Limit Results."
 \(\displaystyle \lim_{x→2}x\)
 \(\displaystyle \lim_{x→2}5\)
Solution
 The limit of \(x\) as \(x\) approaches \(a\) is \(a\): \(\displaystyle \lim_{x→2}x=2\).
 The limit of a constant is that constant: \(\displaystyle \lim_{x→2}5=5\).
We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.
Let \(f(x)\) and \(g(x)\) be defined for all \(x≠a\) over some open interval containing \(a\). Assume that \(L\) and \(M\) are real numbers such that \(\displaystyle \lim_{x→a}f(x)=L\) and \(\displaystyle \lim_{x→a}g(x)=M\). Let \(c\) be a constant. Then, each of the following statements holds:
 Sum law for limits:
\[\displaystyle \lim_{x→a}(f(x)+g(x))=\lim_{x→a}f(x)+\lim_{x→a}g(x)=L+M \nonumber \]
 Difference law for limits:
\[\displaystyle \lim_{x→a}(f(x)−g(x))=\lim_{x→a}f(x)−\lim_{x→a}g(x)=L−M \nonumber \]
 Constant multiple law for limits:
\[\displaystyle \lim_{x→a}cf(x)=c⋅\lim_{x→a}f(x)=cL \nonumber \]
 Product law for limits:
\[\displaystyle \lim_{x→a}(f(x)⋅g(x))=\lim_{x→a}f(x)⋅\lim_{x→a}g(x)=L⋅M \nonumber \]
 Quotient law for limits:
\[\displaystyle \lim_{x→a}\frac{f(x)}{g(x)}=\frac{\displaystyle \lim_{x→a}f(x)}{\displaystyle \lim_{x→a}g(x)}=\frac{L}{M} \nonumber \]
for \(M≠0\).
 Power law for limits:
\[\displaystyle \lim_{x→a}\big(f(x)\big)^n=\big(\lim_{x→a}f(x)\big)^n=L^n \nonumber \]
for every positive integer \(n\).
 Root law for limits:
\[\displaystyle \lim_{x→a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x→a} f(x)}=\sqrt[n]{L} \nonumber \]
for all \(L\) if \(n\) is odd and for \(L≥0\) if \(n\) is even.
We now practice applying these limit laws to evaluate a limit.
Use the limit laws to evaluate \[\lim_{x→−3}(4x+2). \nonumber \]
Solution
Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.
\[\begin{align*} \lim_{x→−3}(4x+2) &= \lim_{x→−3} 4x + \lim_{x→−3} 2 & & \text{Apply the sum law.}\\[4pt] &= 4⋅\lim_{x→−3} x + \lim_{x→−3} 2 & & \text{Apply the constant multiple law.}\\[4pt] &= 4⋅(−3)+2=−10. & & \text{Apply the basic limit results and simplify.} \end{align*}\]
Use the limit laws to evaluate \[\lim_{x→2}\frac{2x^2−3x+1}{x^3+4}. \nonumber \]
Solution
To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.
\[\begin{align*} \lim_{x→2}\frac{2x^2−3x+1}{x^3+4}&=\frac{\displaystyle \lim_{x→2}(2x^2−3x+1)}{\displaystyle \lim_{x→2}(x^3+4)} & & \text{Apply the quotient law, make sure that }(2)^3+4≠0.\\[4pt]
&=\frac{\displaystyle 2⋅\lim_{x→2}x^2−3⋅\lim_{x→2}x+\lim_{x→2}1}{\displaystyle \lim_{x→2}x^3+\lim_{x→2}4} & & \text{Apply the sum law and constant multiple law.}\\[4pt]
&=\frac{\displaystyle 2⋅\left(\lim_{x→2}x\right)^2−3⋅\lim_{x→2}x+\lim_{x→2}1}{\displaystyle \left(\lim_{x→2}x\right)^3+\lim_{x→2}4} & & \text{Apply the power law.}\\[4pt]
&= \frac{2(4)−3(2)+1}{(2)^3+4}=\frac{1}{4}. & & \text{Apply the basic limit laws and simplify.} \end{align*}\]
Use the limit laws to evaluate \(\displaystyle \lim_{x→6}(2x−1)\sqrt{x+4}\). In each step, indicate the limit law applied.
 Hint

Begin by applying the product law.
 Answer

\(11\sqrt{10}\)
Additional Limit Evaluation Techniques
As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for \(\displaystyle \lim_{x→a}f(x)\) to exist when \(f(a)\) is undefined. The following observation allows us to evaluate many limits of this type:
If for all \(x≠a,\;f(x)=g(x)\) over some open interval containing \(a\), then
\[\displaystyle\lim_{x→a}f(x)=\lim_{x→a}g(x). \nonumber \]
To understand this idea better, consider the limit \(\displaystyle \lim_{x→1}\dfrac{x^2−1}{x−1}\).
The function
\[f(x)=\dfrac{x^2−1}{x−1}=\dfrac{(x−1)(x+1)}{x−1}\nonumber \]
and the function \(g(x)=x+1\) are identical for all values of \(x≠1\). The graphs of these two functions are shown in Figure \(\PageIndex{1}\).
We see that
\[\lim_{x→1}\dfrac{x^2−1}{x−1}=\lim_{x→1}\dfrac{(x−1)(x+1)}{x−1}=\lim_{x→1}\,(x+1)=2.\nonumber \]
The limit has the form \(\displaystyle \lim_{x→a}f(x)/g(x)\), where \(\displaystyle\lim_{x→a}f(x)=0\) and \(\displaystyle\lim_{x→a}g(x)=0\). (In this case, we say that \(f(x)/g(x)\) has the indeterminate form \(0/0\).) The following ProblemSolving Strategy provides a general outline for evaluating limits of this type.
 First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.
 We then need to find a function that is equal to \(h(x)=f(x)/g(x)\) for all \(x≠a\) over some interval containing a. To do this, we may need to try one or more of the following steps:
 If \(f(x)\) and \(g(x)\) are polynomials, we should factor each function and cancel out any common factors.
 If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root.
 If \(f(x)/g(x)\) is a complex fraction, we begin by simplifying it.
 Last, we apply the limit laws.
The next examples demonstrate the use of this ProblemSolving Strategy. Example \(\PageIndex{4}\) illustrates the factorandcancel technique; Example \(\PageIndex{5}\) shows multiplying by a conjugate. In Example \(\PageIndex{6}\), we look at simplifying a complex fraction.
Evaluate \(\displaystyle\lim_{x→3}\dfrac{x^2−3x}{2x^2−5x−3}\).
Solution
Step 1. The function \(f(x)=\dfrac{x^2−3x}{2x^2−5x−3}\) is undefined for \(x=3\). In fact, if we substitute 3 into the function we get \(0/0\), which is undefined. Factoring and canceling is a good strategy:
\[\lim_{x→3}\dfrac{x^2−3x}{2x^2−5x−3}=\lim_{x→3}\dfrac{x(x−3)}{(x−3)(2x+1)}\nonumber \]
Step 2. For all \(x≠3,\dfrac{x^2−3x}{2x^2−5x−3}=\dfrac{x}{2x+1}\). Therefore,
\[\lim_{x→3}\dfrac{x(x−3)}{(x−3)(2x+1)}=\lim_{x→3}\dfrac{x}{2x+1}.\nonumber \]
Step 3. Evaluate using the limit laws:
\[\lim_{x→3}\dfrac{x}{2x+1}=\dfrac{3}{7}.\nonumber \]
Evaluate \(\displaystyle \lim_{x→−3}\dfrac{x^2+4x+3}{x^2−9}\).
 Hint

Follow the steps in the ProblemSolving Strategy
 Answer

\(\dfrac{1}{3}\)
Evaluate \( \displaystyle \lim_{x→−1}\dfrac{\sqrt{x+2}−1}{x+1}\).
Solution
Step 1. \( \displaystyle \dfrac{\sqrt{x+2}−1}{x+1}\) has the form \(0/0\) at −1. Let’s begin by multiplying by \(\sqrt{x+2}+1\), the conjugate of \(\sqrt{x+2}−1\), on the numerator and denominator:
\[\lim_{x→−1}\dfrac{\sqrt{x+2}−1}{x+1}=\lim_{x→−1}\dfrac{\sqrt{x+2}−1}{x+1}⋅\dfrac{\sqrt{x+2}+1}{\sqrt{x+2}+1}.\nonumber \]
Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the \((x+1)\) in the denominator cancels out in the end:
\[=\lim_{x→−1}\dfrac{x+1}{(x+1)(\sqrt{x+2}+1)}.\nonumber \]
Step 3. Then we cancel:
\[= \lim_{x→−1}\dfrac{1}{\sqrt{x+2}+1}.\nonumber \]
Step 4. Last, we apply the limit laws:
\[\lim_{x→−1}\dfrac{1}{\sqrt{x+2}+1}=\dfrac{1}{2}.\nonumber \]
Evaluate \( \displaystyle \lim_{x→5}\dfrac{\sqrt{x−1}−2}{x−5}\).
 Hint

Follow the steps in the ProblemSolving Strategy
 Answer

\(\dfrac{1}{4}\)
Evaluate \( \displaystyle \lim_{x→1}\dfrac{\dfrac{1}{x+1}−\dfrac{1}{2}}{x−1}\).
Solution
Step 1. \(\dfrac{\dfrac{1}{x+1}−\dfrac{1}{2}}{x−1}\) has the form \(0/0\) at 1. We simplify the algebraic fraction by multiplying by \(2(x+1)/2(x+1)\):
\[\lim_{x→1}\dfrac{\dfrac{1}{x+1}−\dfrac{1}{2}}{x−1}=\lim_{x→1}\dfrac{\dfrac{1}{x+1}−\dfrac{1}{2}}{x−1}⋅\dfrac{2(x+1)}{2(x+1)}.\nonumber \]
Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor \((x−1)\):
\[=\lim_{x→1}\dfrac{2−(x+1)}{2(x−1)(x+1)}.\nonumber \]
Step 3. Then, we simplify the numerator:
\[=\lim_{x→1}\dfrac{−x+1}{2(x−1)(x+1)}.\nonumber \]
Step 4. Now we factor out −1 from the numerator:
\[=\lim_{x→1}\dfrac{−(x−1)}{2(x−1)(x+1)}.\nonumber \]
Step 5. Then, we cancel the common factors of \((x−1)\):
\[=\lim_{x→1}\dfrac{−1}{2(x+1)}.\nonumber \]
Step 6. Last, we evaluate using the limit laws:
\[\lim_{x→1}\dfrac{−1}{2(x+1)}=−\dfrac{1}{4}.\nonumber \]
Evaluate \( \displaystyle \lim_{x→−3}\dfrac{\dfrac{1}{x+2}+1}{x+3}\).
 Hint

Follow the steps in the ProblemSolving Strategy
 Answer

−1
Example \(\PageIndex{7}\) does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques.
Evaluate \( \displaystyle \lim_{x→0}\left(\dfrac{1}{x}+\dfrac{5}{x(x−5)}\right)\).
Solution:
Both \(1/x\) and \(5/x(x−5)\) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that
\[\dfrac{1}{x}+\dfrac{5}{x(x−5)}=\dfrac{x−5+5}{x(x−5)}=\dfrac{x}{x(x−5)}.\nonumber \]
Thus,
\[\lim_{x→0}\left(\dfrac{1}{x}+\dfrac{5}{x(x−5)}\right)=\lim_{x→0}\dfrac{x}{x(x−5)}=\lim_{x→0}\dfrac{1}{x−5}=−\dfrac{1}{5}.\nonumber \]
Evaluate \( \displaystyle \lim_{x→3}\left(\dfrac{1}{x−3}−\dfrac{4}{x^2−2x−3}\right)\).
 Hint

Use the same technique as Example \(\PageIndex{7}\). Don’t forget to factor \(x^2−2x−3\) before getting a common denominator.
 Answer

\(\dfrac{1}{4}\)
Let’s now revisit onesided limits. Simple modifications in the limit laws allow us to apply them to onesided limits. For example, to apply the limit laws to a limit of the form \(\displaystyle \lim_{x→a^−}h(x)\), we require the function \(h(x)\) to be defined over an open interval of the form \((b,a)\); for a limit of the form \(\displaystyle \lim_{x→a^+}h(x)\), we require the function \(h(x)\) to be defined over an open interval of the form \((a,c)\). Example \(\PageIndex{8A}\) illustrates this point.
Evaluate each of the following limits, if possible.
 \(\displaystyle \lim_{x→3^−}\sqrt{x−3}\)
 \( \displaystyle \lim_{x→3^+}\sqrt{x−3}\)
Solution
Figure \(\PageIndex{2}\) illustrates the function \(f(x)=\sqrt{x−3}\) and aids in our understanding of these limits.
a. The function \(f(x)=\sqrt{x−3}\) is defined over the interval \([3,+∞)\). Since this function is not defined to the left of 3, we cannot apply the limit laws to compute \(\displaystyle\lim_{x→3^−}\sqrt{x−3}\). In fact, since \(f(x)=\sqrt{x−3}\) is undefined to the left of 3, \(\displaystyle\lim_{x→3^−}\sqrt{x−3}\) does not exist.
b. Since \(f(x)=\sqrt{x−3}\) is defined to the right of 3, the limit laws do apply to \(\displaystyle\lim_{x→3^+}\sqrt{x−3}\). By applying these limit laws we obtain \(\displaystyle\lim_{x→3^+}\sqrt{x−3}=0\).
In Example \(\PageIndex{8B}\) we look at onesided limits of a piecewisedefined function and use these limits to draw a conclusion about a twosided limit of the same function.
For \(f(x)=\begin{cases}4x−3, & \mathrm{if} \; x<2 \\ (x−3)^2, & \mathrm{if} \; x≥2\end{cases}\), evaluate each of the following limits:
 \(\displaystyle \lim_{x→2^−}f(x)\)
 \(\displaystyle \lim_{x→2^+}f(x)\)
 \(\displaystyle \lim_{x→2}f(x)\)
Solution
Figure \(\PageIndex{3}\) illustrates the function \(f(x)\) and aids in our understanding of these limits.
a. Since \(f(x)=4x−3\) for all \(x\) in \((−∞,2)\), replace \(f(x)\) in the limit with \(4x−3\) and apply the limit laws:
\[\lim_{x→2^−}f(x)=\lim_{x→2^−}(4x−3)=5\nonumber \]
b. Since \(f(x)=(x−3)^2\)for all \(x\) in \((2,+∞)\), replace \(f(x)\) in the limit with \((x−3)^2\) and apply the limit laws:
\[\lim_{x→2^+}f(x)=\lim_{x→2^−}(x−3)^2=1. \nonumber \]
c. Since \(\displaystyle \lim_{x→2^−}f(x)=5\) and \(\displaystyle \lim_{x→2^+}f(x)=1\), we conclude that \(\displaystyle \lim_{x→2}f(x)\) does not exist.
Graph \(f(x)=\begin{cases}−x−2, & \mathrm{if} \; x<−1\\ 2, & \mathrm{if} \; x=−1 \\ x^3, & \mathrm{if} \; x>−1\end{cases}\) and evaluate \(\displaystyle \lim_{x→−1^−}f(x)\).
 Hint

Use the method in Example \(\PageIndex{8B}\) to evaluate the limit.
 Answer

\[\lim_{x→−1^−}f(x)=−1\nonumber \]
We now turn our attention to evaluating a limit of the form \(\displaystyle \lim_{x→a}\dfrac{f(x)}{g(x)}\), where \(\displaystyle \lim_{x→a}f(x)=K\), where \(K≠0\) and \(\displaystyle \lim_{x→a}g(x)=0\). That is, \(f(x)/g(x)\) has the form \(K/0,K≠0\) at \(a\).
Evaluate \(\displaystyle \lim_{x→2^−}\dfrac{x−3}{x^2−2x}\).
Solution
Step 1. After substituting in \(x=2\), we see that this limit has the form \(−1/0\). That is, as \(x\) approaches \(2\) from the left, the numerator approaches \(−1\); and the denominator approaches \(0\). Consequently, the magnitude of \(\dfrac{x−3}{x(x−2)} \) becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:
\[\lim_{x→2^−}\dfrac{x−3}{x^2−2x}=\lim_{x→2^−}\dfrac{x−3}{x(x−2)} \nonumber \]
Step 2. Since \(x−2\) is the only part of the denominator that is zero when 2 is substituted, we then separate \(1/(x−2)\) from the rest of the function:
\[=\lim_{x→2^−}\dfrac{x−3}{x}⋅\dfrac{1}{x−2} \nonumber \]
Step 3. Using the Limit Laws, we can write:
\[=\left(\lim_{x→2^−}\dfrac{x−3}{x}\right)\cdot\left(\lim_{x→2^−}\dfrac{1}{x−2}\right). \nonumber \]
Step 4. \(\displaystyle \lim_{x→2^−}\dfrac{x−3}{x}=−\dfrac{1}{2}\) and \(\displaystyle \lim_{x→2^−}\dfrac{1}{x−2}=−∞\). Therefore, the product of \((x−3)/x\) and \(1/(x−2)\) has a limit of \(+∞\):
\[\lim_{x→2^−}\dfrac{x−3}{x^2−2x}=+∞. \nonumber \]
Evaluate \(\displaystyle \lim_{x→1}\dfrac{x+2}{(x−1)^2}\).
 Solution

Use the methods from Example \(\PageIndex{9}\).
 Answer

\(+∞\)
The Squeeze Theorem
The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point \(a\) that is unknown, between two functions having a common known limit at \(a\). Figure \(\PageIndex{4}\) illustrates this idea.
Let \(f(x),g(x)\), and \(h(x)\) be defined for all \(x≠a\) over an open interval containing \(a\). If
\[f(x)≤g(x)≤h(x) \nonumber \]
for all \(x≠a\) in an open interval containing \(a\) and
\[\lim_{x→a}f(x)=L=\lim_{x→a}h(x) \nonumber \]
where \(L\) is a real number, then \(\displaystyle \lim_{x→a}g(x)=L.\)
Apply the squeeze theorem to evaluate \(\displaystyle \lim_{x→0} x \cos x\).
Solution
Because \(−1≤\cos x≤1\) for all \(x\), we have \(−x≤x \cos x≤x\) for \(x≥0\) and \(−x≥x \cos x ≥ x\) for \(x≤0\) (if \(x\) is negative the direction of the inequalities changes when we multiply). Since \(\displaystyle \lim_{x→0}(−x)=0=\lim_{x→0}x\), from the squeeze theorem, we obtain \(\displaystyle \lim_{x→0}x \cos x=0\). The graphs of \(f(x)=−x,\;g(x)=x\cos x\), and \(h(x)=x\) are shown in Figure \(\PageIndex{5}\).
Use the squeeze theorem to evaluate \(\displaystyle \lim_{x→0}x^2 \sin\dfrac{1}{x}\).
 Hint

Use the fact that \(−x^2≤x^2\sin (1/x) ≤ x^2\) to help you find two functions such that \(x^2\sin (1/x)\) is squeezed between them.
 Answer

0
Limits of Polynomial and Rational Functions
By now you have probably noticed that, in each of the previous examples, it has been the case that \(\displaystyle \lim_{x→a}f(x)=f(a)\). This is not always true, but it does hold for all polynomials for any choice of \(a\) and for all rational functions at all values of \(a\) for which the rational function is defined.
Limits of Polynomial and Rational Functions
Let \(p(x)\) and \(q(x)\) be polynomial functions. Let \(a\) be a real number. Then,
\[\lim_{x→a}p(x)=p(a) \nonumber \]
\[\lim_{x→a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)} \nonumber \]
when \(q(a)≠0\).
To see that this theorem holds, consider the polynomial
\[p(x)=c_nx^n+c_{n−1}x^{n−1}+⋯+c_1x+c_0. \nonumber \]
By applying the sum, constant multiple, and power laws, we end up with
\[ \begin{align*} \lim_{x→a}p(x) &= \lim_{x→a}(c_nx^n+c_{n−1}x^{n−1}+⋯+c_1x+c_0) \\[4pt] &= c_n\left(\lim_{x→a}x\right)^n+c_{n−1}\left(\lim_{x→a}x\right)^{n−1}+⋯+c_1\left(\lim_{x→a}x\right)+\lim_{x→a}c_0 \\[4pt] &= c_na^n+c_{n−1}a^{n−1}+⋯+c_1a+c_0 \\[4pt] &= p(a) \end{align*}\]
It now follows from the quotient law that if \(p(x)\) and \(q(x)\) are polynomials for which \(q(a)≠0\),
then
\[\lim_{x→a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}. \nonumber \]
Example \(\PageIndex{6}\): Evaluating a Limit of a Rational Function
Evaluate the \(\displaystyle \lim_{x→3}\frac{2x^2−3x+1}{5x+4}\).
Solution
Since 3 is in the domain of the rational function \(f(x)=\displaystyle \frac{2x^2−3x+1}{5x+4}\), we can calculate the limit by substituting 3 for \(x\) into the function. Thus,
\[\lim_{x→3}\frac{2x^2−3x+1}{5x+4}=\frac{10}{19}. \nonumber \]
Exercise \(\PageIndex{6}\)
Evaluate \(\displaystyle \lim_{x→−2}(3x^3−2x+7)\).
Use LIMITS OF POLYNOMIAL AND RATIONAL FUNCTIONS as reference
−13