Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

1.3: Limit calculations for algebraic expressions

  • Page ID
    10246
  • [ "stage:draft", "article:topic" ]

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    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 \(c\),

    1. \(\displaystyle \lim_{x→a}x=a\)
    2. \(\displaystyle \lim_{x→a}c=c\)
    3. \(\displaystyle \lim_{x→0^+} \dfrac{1}{x}=\infty\)
    4. \(\displaystyle \lim_{x→0^-} \dfrac{1}{x}= -\infty\)

    Example \(\PageIndex{1}\): Evaluating Using Basic Results

    Evaluate each of the following limits:

    1. \(\displaystyle \lim_{x→2}x\)
    2. \(\displaystyle \lim_{x→2}5\)

    Solution:

    1. The limit of x as x approaches a is a: \(\displaystyle \lim_{x→2}x=2\).
    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 Problem-Solving Strategy provides a general outline for evaluating limits of this type.

    Problem-Solving Strategy: Calculating a Limit When \(f(x)/g(x)\)has the Indeterminate Form 0/0

    1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.
    2. 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:
      1. If \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are polynomials, we should factor each function and cancel out any common factors.
      2. 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.
      3. If \(\displaystyle f(x)/g(x)\) is a complex fraction, we begin by simplifying it.
    3. Last, we apply the limit laws.

    The next examples demonstrate the use of this Problem-Solving Strategy. Example illustrates the factor-and-cancel 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 Problem-Solving 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 Problem-Solving Strategy 

    Answer

    \(\displaystyle \frac{1}{4}\)

    Example \(\PageIndex{6}\): Evaluating a Limit by Simplifying a Complex Fraction

    Evaluate \(\displaystyle \lim_{x→1}\frac{\frac{1}{x+1}−\frac{1}{2}}{x−1}\).
     
    Solution

    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 Problem-Solving 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}(\frac{1}{x}+\frac{5}{x(x−5)}).

    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 one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided 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 One-Sided Limit Using the Limit Laws

    Evaluate each of the following limits, if possible.

    1. \(\displaystyle \lim_{x→3−}\sqrt{x−3}\)
    2. \(\displaystyle \lim_{x→3+}\sqrt{x−3}\)

    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 one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.

    Example \(\PageIndex{9}\): Evaluating a Two-Sided 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:

    1. \(\displaystyle \lim_{x→2−}f(x)\)
    2. \(\displaystyle \lim_{x→2+}f(x)\)

    Contributors

    • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

    • Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)