1.3: Limit calculations for algebraic expressions
 Page ID
 10246
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Basic Results
The following basic results, together with the other properties of limits, allow us to evaluate limits of many algebraic functions.
Basic Results
For any real number \(a\) and any constant \(c\),
 \(\displaystyle \lim_{x→a}x=a\)
 \(\displaystyle \lim_{x→a}c=c\)
 \(\displaystyle \lim_{x→0^+} \dfrac{1}{x}=\infty\)
 \(\displaystyle \lim_{x→0^} \dfrac{1}{x}= \infty\)
Example \(\PageIndex{1}\): Evaluating Using Basic Results
Evaluate each of the following limits:
 \(\displaystyle \lim_{x→2}x\)
 \(\displaystyle \lim_{x→2}5\)
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 individual properties of limits. The proofs that these properties hold are omitted here.
Properties of Limits
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 \(\lim_{x→a}f(x)=L\) and \(\lim_{x→a}g(x)=M\). 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: \(\displaystyle \lim_{x→a}\frac{f(x)}{g(x)}=\frac{\lim_{x→a}f(x)}{\lim_{x→a}g(x)}=\dfrac{L}{M}\) for M≠0 
Power law for limits: \(\displaystyle \lim_{x→a}(f(x))^n=(\lim_{x→a}f(x))^n=L^n\) for every positive integer n. 
Root law for limits: \(\displaystyle \lim_{x→a}\sqrt[n]{f(x)}=\lim_{x→a}\sqrt[n]{f(x)}=\sqrt[n]{L}\) for all L if n is odd and for \(L≥0\) if n is even. 
We now practice applying these properties to evaluate a limit.
Example \(\PageIndex{2A}\): Evaluating Limit
Evaluate \[\lim_{x→−3}(4x+2).\]
Solution
Let’s apply the properties of limits 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 property, the new limits must exist for the property to be applied.
\(\displaystyle \lim_{x→−3}(4x+2) = \lim_{x→−3} 4x + \lim_{x→−3} 2\) Apply the sum.
\(\displaystyle = 4⋅\lim_{x→−3} x + \lim_{x→−3} 2\) Apply the constant multiple.
\(\displaystyle = 4⋅(−3)+2=−10.\) Apply the basic limit results and simplify.
Example \(\PageIndex{2B}\): Using Properties Repeatedly
Evaluate \[\lim_{x→2}\frac{2x^2−3x+1}{x^3+4}.\]
Solution
To find this limit, we need to apply the properties 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 property to be applied.
\(\displaystyle\lim_{x→2}\frac{2x^2−3x+1}{x^3+4}=\frac{\lim_{x→2}(2x^2−3x+1)}{\lim_{x→2}(x^3+4)}\) Apply the quotient law, make sure that \(\displaystyle(2)^3+4≠0.\)
\(\displaystyle =\frac{2⋅\lim_{x→2}x^2−3⋅\lim_{x→2}x+\lim_{x→2}1}{\lim_{x→2}x^3+\lim_{x→2}4}\) Apply the sum law and constant multiple law
\(\displaystyle =\frac{2⋅(\lim_{x→2}x)^2−3⋅\lim_{x→2}x+\lim_{x→2}1}{(\lim_{x→2}x)^3+\lim_{x→2}4}\) Apply the power law.
\(\displaystyle =\frac{2(4)−3(2)+1}{(2)^3+4}=\frac{1}{4}\). Apply the basic limit laws and simplify.
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 \(\lim_{x→a}f(x)=f(a)\). 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 \(\displaystyle p(x)\) 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 \(\displaystyle q(a)≠0\).
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
\(\displaystyle \lim_{x→a}p(x)\) = \(\lim_{x→a}(c_nx^n+c_{n−1}x^{n−1}+⋯+c_1x+c_0)\)
\(\displaystyle =c_n(\lim_{x→a}x)^n+c_{n−1}(\lim_{x→a}x)^{n−1}+⋯+c_1(\lim_{x→a}x)+\lim_{x→a}c_0\)
\(\displaystyle =c_na^n+c_{n−1}a^{n−1}+⋯+c_1a+c_0\)
\(\displaystyle =p(a)\).
It now follows from the quotient law that if \(\displaystyle p(x)\) and \(\displaystyle q(x)\) are polynomials for which \(\displaystyle q(a)≠0\),
then
\(\displaystyle \lim_{x→a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}\).
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)=\frac{2x^2−3x+1}{5x+4}, we can calculate the limit by substituting 3 for x into the function. Thus,
\(\displaystyle \lim_{x→3}\frac{2x^2−3x+1}{5x+4}=\frac{10}{19}\).
Exercise \(\PageIndex{2}\)
Evaluate \(\displaystyle \lim_{x→−2}(3x^3−2x+7)\).
Hint

Use LIMITS OF POLYNOMIAL AND RATIONAL FUNCTIONS as reference
 Answer

−13
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 \(\lim_{x→a}f(x)\) 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 \(\lim_{x→a}f(x)=\lim_{x→a}g(x)\).
The limit has the form limx→af(x)g(x), where \(\lim_{x→a}f(x)=0 and \lim_{x→a}g(x)=0\). (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 \(\displaystyle h(x)=f(x)/g(x)\) for all \(\displaystyle x≠a\) over some interval containing a. To do this, we may need to try one or more of the following steps:
 If \(\displaystyle f(x)\) and \(\displaystyle g(x)\) 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 \(\displaystyle f(x)/g(x)\) 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 illustrates the factorandcancel technique; Example shows multiplying by a conjugate. In Example, we look at simplifying a complex fraction.
Example \(\PageIndex{4}\): Evaluating a Limit by Factoring and Canceling
Evaluate \(\displaystyle \lim_{x→3}\frac{x^2−3x}{2x^2−5x−3}\).
Solution
Step 1. The function \(\displaystyle f(x)=\frac{x^2−3x}{2x^2−5x−3}\) is undefined for \(\displaystyle x=3\). In fact, if we substitute 3 into the function we get \(\displaystyle 0/0\), which is undefined. Factoring and canceling is a good strategy:
\(\displaystyle \lim_{x→3}\frac{x^2−3x}{2x^2−5x−3}=\lim_{x→3}\frac{x(x−3)}{(x−3)(2x+1)}\)
Step 2. For all \(\displaystyle x≠3,\frac{x^2−3x}{2x^2−5x−3}=\frac{x}{2x+1}\). Therefore,
\(\displaystyle \lim_{x→3}\frac{x(x−3)}{(x−3)(2x+1)}=\lim_{x→3}\frac{x}{2x+1}\).
Step 3. Evaluate using the limit laws:
\(\displaystyle \lim_{x→3}\frac{x}{2x+1}=\frac{3}{7}\).
Exercise \(\PageIndex{3}\)
Evaluate \(\displaystyle \lim_{x→−3}\frac{x^2+4x+3}{x^2−9}\).
 Hint

Follow the steps in the ProblemSolving Strategy
 Answer

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

Follow the steps in the ProblemSolving Strategy
 Answer

\(\displaystyle \frac{1}{4}\)
Example \(\PageIndex{6}\): Evaluating a Limit by Simplifying a Complex Fraction
Step 1. \(\displaystyle \frac{\frac{1}{x+1}−\frac{1}{2}}{x−1}\) has the form \(\displaystyle 0/0\) at 1. We simplify the algebraic fraction by multiplying by \(\displaystyle 2(x+1)/2(x+1)\):
\(\displaystyle \lim_{x→1}\frac{\frac{1}{x+1}−\frac{1}{2}}{x−1}=\lim_{x→1}\frac{\frac{1}{x+1}−\frac{1}{2}}{x−1}⋅\frac{2(x+1)}{2(x+1)}\).
Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor \((x−1)\):
\(\displaystyle =\lim_{x→1}\frac{2−(x+1)}{2(x−1)(x+1)}\).
Step 3. Then, we simplify the numerator:
\(\displaystyle =\lim_{x→1}\frac{−x+1}{2(x−1)(x+1)}\).
Step 4. Now we factor out −1 from the numerator:
\(\displaystyle =\lim_{x→1}\frac{−(x−1)}{2(x−1)(x+1)}\).
Step 5. Then, we cancel the common factors of \((x−1)\):
\(\displaystyle =\lim_{x→1}\frac{−1}{2(x+1)}\).
Step 6. Last, we evaluate using the limit laws:
\(\displaystyle \lim_{x→1}\frac{−1}{2(x+1})=−\frac{1}{4}\).
Exercise \(\PageIndex{5}\)
Evaluate \(\displaystyle \lim_{x→−3}\frac{\frac{1}{x+2}+1}{x+3}\).
 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
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
\(\displaystyle \frac{1}{x}+\frac{5}{x(x−5)}=\frac{x−5+5}{x(x−5)}=\frac{x}{x(x−5)}\).
Thus,
\(\displaystyle \lim_{x→0}(\frac{1}{x}+\frac{5}{x(x−5)})=\lim_{x→0}\frac{x}{x(x−5)}=\lim_{x→0}\frac{1}{x−5}=−\frac{1}{5}\)
Exercise \(\PageIndex{6}\)
Evaluate \(\displaystyle \lim_{x→3}(\frac{1}{x−3}−\frac{4}{x^2−2x−3}).
 Hint

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

\(\frac{1}{4}\)
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 \(\lim_{x→a−}h(x)\), we require the function \(h(x)\) to be defined over an open interval of the form \((b,a)\); for a limit of the form \(\lim_{x→a+}h(x)\), we require the function \(h(x)\) to be defined over an open interval of the form \((a,c)\). Example illustrates this point.
Example \(\PageIndex{8}\): 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.
Figure \(\PageIndex{2}\): The graph shows the function \(f(x)=\sqrt{x−3}\).
a. The function \(\displaystyle f(x)=\sqrt{x−3}\) is defined over the interval \(\displaystyle [3,+∞)\). 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 \(\displaystyle f(x)=\sqrt{x−3}\) is undefined to the left of 3, \(\displaystyle \lim_{x→3−}\sqrt{x−3}\) does not exist.
b. Since \(\displaystyle f(x)=\sqrt{x−3}\) 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 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 \(\PageIndex{9}\): Evaluating a TwoSided Limit Using the Limit Laws
For \(\displaystyle f(x)=\begin{cases}4x−3 & if x<2 \\ (x−3)^2 & if x≥2\end{cases}\), evaluate each of the following limits:
 \(\displaystyle \lim_{x→2−}f(x)\)
 \(\displaystyle \lim_{x→2+}f(x)\)
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)