1.3: Limit calculations for algebraic expressions
 Page ID
 10246
Learning Objectives
 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.
In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits.
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 the limits of many algebraic functions.
Basic Limit Results
For any real number \(a\) and any constant \(c\),
 \(\displaystyle \lim_{x→a}x=a\)
 \(\displaystyle \lim_{x→a}c=c\)
Example \(\PageIndex{1}\): Evaluating a Basic Limit
Evaluate each of the following limits using Note.
 \(\displaystyle \lim_{x→2}x\)
 \(\displaystyle \lim_{x→2}5\)
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.
Limit Laws
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:
\[\displaystyle \lim_{x→a}(f(x)+g(x))=\lim_{x→a}f(x)+\lim_{x→a}g(x)=L+M\]
 Difference law for limits:
\[\displaystyle \lim_{x→a}(f(x)−g(x))=\lim_{x→a}f(x)−\lim_{x→a}g(x)=L−M\]
 Constant multiple law for limits:
\[\displaystyle \lim_{x→a}cf(x)=c⋅\lim_{x→a}f(x)=cL\]
 Product law for limits:
\[\displaystyle \lim_{x→a}(f(x)⋅g(x))=\lim_{x→a}f(x)⋅\lim_{x→a}g(x)=L⋅M\]
 Quotient law for limits:
\[\displaystyle \lim_{x→a}\frac{f(x)}{g(x)}=\frac{\displaystyle \lim_{x→a}f(x)}{\displaystyle \lim_{x→a}g(x)}=\frac{L}{M}\]
for \(M≠0\).
 Power law for limits:
\[\displaystyle \lim_{x→a}\big(f(x)\big)^n=\big(\lim_{x→a}f(x)\big)^n=L^n\]
for every positive integer \(n\).
 Root law for limits:
\[\displaystyle \lim_{x→a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x→a} f(x)}=\sqrt[n]{L}\]
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.
Example \(\PageIndex{2A}\): Evaluating a Limit Using Limit Laws
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.
\[\begin{align*} \lim_{x→−3}(4x+2) &= \lim_{x→−3} 4x + \lim_{x→−3} 2 & & \text{Apply the sum law.}\\[4pt] &= 4⋅\lim_{x→−3} x + \lim_{x→−3} 2 & & \text{Apply the constant multiple law.}\\[4pt] &= 4⋅(−3)+2=−10. & & \text{Apply the basic limit results and simplify.} \end{align*}\]
Example \(\PageIndex{2B}\): Using Limit Laws Repeatedly
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.
\[\begin{align*} \lim_{x→2}\frac{2x^2−3x+1}{x^3+4}&=\frac{\displaystyle \lim_{x→2}(2x^2−3x+1)}{\displaystyle \lim_{x→2}(x^3+4)} & & \text{Apply the quotient law, make sure that }(2)^3+4≠0.\\[4pt]
&=\frac{\displaystyle 2⋅\lim_{x→2}x^2−3⋅\lim_{x→2}x+\lim_{x→2}1}{\displaystyle \lim_{x→2}x^3+\lim_{x→2}4} & & \text{Apply the sum law and constant multiple law.}\\[4pt]
&=\frac{\displaystyle 2⋅\left(\lim_{x→2}x\right)^2−3⋅\lim_{x→2}x+\lim_{x→2}1}{\displaystyle \left(\lim_{x→2}x\right)^3+\lim_{x→2}4} & & \text{Apply the power law.}\\[4pt]
&= \frac{2(4)−3(2)+1}{(2)^3+4}=\frac{1}{4}. & & \text{Apply the basic limit laws and simplify.} \end{align*}\]
Exercise \(\PageIndex{2}\)
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
\[\displaystyle\lim_{x→a}f(x)=\lim_{x→a}g(x).\]
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.
ProblemSolving Strategy: Calculating a Limit When \(f(x)/g(x)\) has the Indeterminate Form \(0/0\)
 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.
Example \(\PageIndex{4}\): Evaluating a Limit by Factoring and Canceling
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:
\[\lim_{x→3}\dfrac{x}{2x+1}=\dfrac{3}{7}.\nonumber\]
Exercise \(\PageIndex{4}\)
Evaluate \(\displaystyle \lim_{x→−3}\dfrac{x^2+4x+3}{x^2−9}\).
 Hint

Follow the steps in the ProblemSolving Strategy
 Answer

\(\dfrac{1}{3}\)
Example \(\PageIndex{5}\): Evaluating a Limit by Multiplying by a Conjugate
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:
\[\lim_{x→−1}\dfrac{1}{\sqrt{x+2}+1}=\dfrac{1}{2}.\nonumber\]
Exercise \(\PageIndex{5}\)
Evaluate \( \displaystyle \lim_{x→5}\dfrac{\sqrt{x−1}−2}{x−5}\).
 Hint

Follow the steps in the ProblemSolving Strategy
 Answer

\(\dfrac{1}{4}\)
Example \(\PageIndex{6}\): Evaluating a Limit by Simplifying a Complex Fraction
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:
\[\lim_{x→1}\dfrac{−1}{2(x+1)}=−\dfrac{1}{4}.\nonumber\]
Exercise \(\PageIndex{6}\)
Evaluate \( \displaystyle \lim_{x→−3}\dfrac{\dfrac{1}{x+2}+1}{x+3}\).
 Hint

Follow the steps in the ProblemSolving Strategy and
 Answer

−1
Example 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.
Example \(\PageIndex{7}\): Evaluating a Limit When the Limit Laws Do Not Apply
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,
\[\lim_{x→0}\left(\dfrac{1}{x}+\dfrac{5}{x(x−5)}\right)=\lim_{x→0}\dfrac{x}{x(x−5)}=\lim_{x→0}\dfrac{1}{x−5}=−\dfrac{1}{5}.\nonumber\]
Exercise \(\PageIndex{7}\)
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.
Example \(\PageIndex{8A}\): Evaluating a OneSided Limit Using the Limit Laws
Evaluate each of the following limits, if possible.
 \(\displaystyle \lim_{x→3^−}\sqrt{x−3}\)
 \( \displaystyle \lim_{x→3^+}\sqrt{x−3}\)
Solution
Figure 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.
Example \(\PageIndex{8B}\): Evaluating a TwoSided Limit Using the Limit Laws
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 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.
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.
Example \(\PageIndex{9}\): Evaluating a Limit of the Form \(K/0,K≠0\) Using the Limit Laws
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 \(+∞\):
\[\lim_{x→2^−}\dfrac{x−3}{x^2−2x}=+∞. \nonumber\]
Exercise \(\PageIndex{9}\)
Evaluate \(\displaystyle \lim_{x→1}\dfrac{x+2}{(x−1)^2}\).
 Solution

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

\(+∞\)
Example \(\PageIndex{10}\)
Evaluate \(\displaystyle \lim_{x→4} \dfrac{(x3)}{(4x)(x+3)}\).
Solution
If we substitute \(x=4\) into \(\dfrac{(x3)}{(4x)(x+3)}\), we get \( \dfrac{1}{0}\). In this case we use the following sign chart to decide \(\displaystyle \lim_{x→4} \dfrac{(x3)}{(4x)(x+3)}\). Notice that the point(s) where the denominator zero is \(x=4\) and \(x=3\), and the point(s) where the numerator zero is \(x=3\).
\((\infty, 3)\)  \((3,3)\)  \((3,4)\)  \((4, \infty)\)  
\((x3)\)  \(\)  \(\) 

\(+\)  
\((4x)\)  \(+\) 


\( \)  
\((x+3)\)  \(\) 


\(+\)  
+   
Thus \(\displaystyle \lim_{x→4^} \dfrac{(x3)}{(4x)(x+3)} = \infty\), and \(\displaystyle \lim_{x→4^+} \dfrac{(x3)}{(4x)(x+3)}=\infty\). Therefore \(\displaystyle \lim_{x→4} \dfrac{(x3)}{(4x)(x+3)}=DNE.\)
Exercise \(\PageIndex{10}\)
Evaluate \(\displaystyle \lim_{x→3} \dfrac{(x+3)}{(4x)(x3)}\).
 Answer

Add texts here. Do not delete this text first.
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.
Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)
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,
\[\lim_{x→a}p(x)=p(a)\]
\[\lim_{x→a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}\]
when \(q(a)≠0\).
To see that this theorem holds, consider the polynomial
\[p(x)=c_nx^n+c_{n−1}x^{n−1}+⋯+c_1x+c_0.\]
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
\[\lim_{x→a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}.\]
Example \(\PageIndex{3}\): 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 \(\PageIndex{3}\)
Evaluate \(\displaystyle \lim_{x→−2}(3x^3−2x+7)\).
Use LIMITS OF POLYNOMIAL AND RATIONAL FUNCTIONS as reference
−13