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Mathematics LibreTexts

1.3: Limit calculations for algebraic expressions

  • Page ID
    10246
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    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 \(c\),

    1. \(\displaystyle \lim_{x→a}x=a\)
    2. \(\displaystyle \lim_{x→a}c=c\)

    Example \(\PageIndex{1}\): Evaluating a Basic Limit

    Evaluate each of the following limits using Note.

    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 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

    \(11\sqrt{10}\)

    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 \(\displaystyle \lim_{x→−2}(3x^3−2x+7)\).

    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 \(\displaystyle \lim_{x→a}f(x)\) 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 \(\PageIndex{1}\).

    Two graphs side by side. The first is a graph of g(x) = x + 1, a linear function with y intercept at (0,1) and x intercept at (-1,0). The second is a graph of f(x) = (x^2 – 1) / (x – 1). This graph is identical to the first for all x not equal to 1, as there is an open circle at (1,2) in the second graph.
    Figure \(\PageIndex{1}\): The graphs of \(f(x)\) and \(g(x)\) are identical for all \(x≠1\). Their limits at 1 are equal.

    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 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 \(h(x)=f(x)/g(x)\) for all \(x≠a\) over some interval containing a. To do this, we may need to try one or more of the following steps:
      1. If \(f(x)\) and \(g(x)\) 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 \(f(x)/g(x)\) 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 \(\PageIndex{4}\) illustrates the factor-and-cancel 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 Problem-Solving 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 Problem-Solving 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 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 \( \displaystyle \lim_{x→0}\left(\dfrac{1}{x}+\dfrac{5}{x(x−5)}\right)\).

    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 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 \(\displaystyle \lim_{x→a^−}h(x)\), 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 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.

    A graph of the function f(x) = sqrt(x-3). Visually, the function looks like the top half of a parabola opening to the right with vertex at (3,0).
    Figure \(\PageIndex{2}\): The graph shows the function \(f(x)=\sqrt{x−3}\).

    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 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 \(\PageIndex{8B}\): Evaluating a Two-Sided 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:

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

    Solution

    Figure illustrates the function \(f(x)\) and aids in our understanding of these limits.

    The graph of a piecewise function with two segments. For x<2, the function is linear with the equation 4x-3. There is an open circle at (2,5). The second segment is a parabola and exists for x>=2, with the equation (x-3)^2. There is a closed circle at (2,1). The vertex of the parabola is at (3,0).
    Figure \(\PageIndex{3}\): This graph shows a function \(f(x)\).

    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 \(\displaystyle \lim_{x→4} \dfrac{(x-3)}{(4-x)(x+3)}\).

    Solution

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

     

      \((-\infty, -3)\) \((-3,3)\) \((3,4)\) \((4, \infty)\)
    \((x-3)\) \(-\) \(-\)
    \(+\)
    \(+\)
    \((4-x)\) \(+\)
    \(+\)
    \(+\)
    \(- \)
    \((x+3)\) \(-\)
    \(+\)
    \(+\)
    \(+\)
          + -

     

    Thus   \(\displaystyle \lim_{x→4^-} \dfrac{(x-3)}{(4-x)(x+3)} = \infty\), and  \(\displaystyle \lim_{x→4^+} \dfrac{(x-3)}{(4-x)(x+3)}=-\infty\). Therefore  \(\displaystyle \lim_{x→4} \dfrac{(x-3)}{(4-x)(x+3)}=DNE.\)

    Exercise \(\PageIndex{10}\)

    Evaluate \(\displaystyle \lim_{x→3} \dfrac{(x+3)}{(4-x)(x-3)}\).

    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 CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

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