{"id":198,"date":"2023-09-20T22:48:13","date_gmt":"2023-09-20T22:48:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/more-limit-evaluation-techniques\/"},"modified":"2024-08-05T12:33:17","modified_gmt":"2024-08-05T12:33:17","slug":"the-limit-laws-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-limit-laws-learn-it-2\/","title":{"raw":"The Limit Laws: Learn It 2","rendered":"The Limit Laws: Learn It 2"},"content":{"raw":"

Evaluating Limits Cont.<\/h2>\r\n

More Limit Evaluation Techniques<\/h3>\r\n

Calculating limits is straightforward for polynomials and some rational functions using direct substitution. However, when direct substitution results in an undefined form, such as [latex]\\frac{0}{0}[\/latex], further techniques are needed.<\/p>\r\n

\r\n

Consider the function [latex]\\frac{x^2-1}{x-1}[\/latex].<\/p>\r\n

Direct substitution for [latex]x=1[\/latex] would yield the indeterminate form [latex]\\frac{0}{0}[\/latex].<\/p>\r\n

By factoring [latex]f(x)[\/latex] to its simplest form,<\/p>\r\n

[latex]\\begin{array}{ccc}\\hfill f(x)& =\\dfrac{x^2-1}{x-1}\\hfill \\\\ & =\\dfrac{(x-1)(x+1)}{x-1}\\hfill \\\\ & = x+1\\hfill \\end{array}[\/latex]<\/center>\r\n

we see that except for [latex]x=1,[\/latex] where the function is not defined, the limit as [latex]x[\/latex] approaches [latex]1[\/latex] is the same as the value of [latex]g(x)=x+1[\/latex] at that point, which is [latex]2[\/latex].<\/p>\r\n

[latex]\\begin{array}{cc}\\hfill \\underset{x\\to 1}{\\lim}\\dfrac{x^2-1}{x-1}& =\\underset{x\\to 1}{\\lim}\\dfrac{(x-1)(x+1)}{x-1}\\hfill \\\\ & =\\underset{x\\to 1}{\\lim}(x+1)\\hfill \\\\ & =2\\hfill \\end{array}[\/latex]<\/center>The graphs of these two functions are shown in Figure 1.\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"Two Figure 1. The graphs of [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are identical for all [latex]x\\ne 1[\/latex]. Their limits at 1 are equal.[\/caption]\r\n<\/section>\r\n

This shows how simplification can reveal a continuous limit where direct substitution fails.<\/p>\r\n

When the function assumes the form [latex]\\frac{0}{0}[\/latex] upon direct substitution, it's an indeterminate form<\/strong>.<\/p>\r\n

\r\n

indeterminate form in limits<\/strong><\/h3>\r\n

When direct substitution in a function yields [latex]\\frac{0}{0}[\/latex], it indicates an indeterminate form requiring further analysis to calculate the limit.<\/p>\r\n<\/section>\r\n

\r\n

How To: Solve Indeterminate [latex]\\frac{0}{0}[\/latex] Limits<\/strong><\/p>\r\n

    \r\n\t
  1. Verify the function has the appropriate form and cannot be evaluated immediately using the limit laws.<\/li>\r\n\t
  2. Find an equivalent function [latex]h(x)=f(x)\/g(x)[\/latex] valid for all [latex]x[\/latex] near the point of interest, except at the point itself. To do this, we may need to try one or more of the following steps:
    \r\n
      \r\n\t
    1. Factor and simplify polynomials to cancel common terms.<\/li>\r\n\t
    2. For square roots, use conjugates to rationalize.<\/li>\r\n\t
    3. Simplify complex fractions.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t
    4. Apply the limit laws to the simplified expression to find the limit.<\/li>\r\n<\/ol>\r\n<\/section>\r\n

      Evaluate a Limit by Factoring and Simplifying Polynomials<\/h3>\r\n

      A key step in resolving indeterminate limits involves factoring and simplifying polynomials to cancel out common terms. By extracting the GCF, applying special product formulas for polynomials, and using grouping techniques for trinomials, we can often simplify the expressions and resolve the indeterminate forms.<\/p>\r\n

      \r\n

      How To: Evaluate a Limit by Factoring and Simplifying Polynomials<\/strong><\/p>\r\n

        \r\n\t
      1. Factor the GCF out from the polynomial terms to reduce complexity.<\/li>\r\n\t
      2. Apply factoring rules for difference of squares, perfect square trinomials, and sum\/difference of cubes to break down terms further.<\/li>\r\n\t
      3. For trinomials, utilize grouping for efficient factorization.<\/li>\r\n\t
      4. After simplification, re-evaluate the limit with the new, simplified expression.<\/li>\r\n<\/ol>\r\n<\/section>\r\n
        \r\n

        [latex]\\begin{array}{ll} \\text{difference of squares} & a^2 - b^2 = (a+b)(a-b) \\\\ \\text{perfect square trinomial} & a^2 + 2ab + b^2 = (a+b)^2 \\\\ \\text{sum of cubes} & a^3 + b^3 = (a+b)(a^2 - ab + b^2) \\\\ \\text{difference of cubes} & a^3 - b^3 = (a-b)(a^2 + ab + b^2) \\end{array}[\/latex]<\/p>\r\n<\/section>\r\n

        \r\n

        Evaluate [latex]\\underset{x\\to 3}{\\lim}\\dfrac{x^2-3x}{2x^2-5x-3}[\/latex]<\/p>\r\n

        [reveal-answer q=\"fs-id1170572335183\"]Show Solution[\/reveal-answer]
        \r\n[hidden-answer a=\"fs-id1170572335183\"]<\/p>\r\n

        The function [latex]f(x)=\\frac{x^2-3x}{2x^2-5x-3}[\/latex] is undefined for [latex]x=3[\/latex]. In fact, if we substitute [latex]3[\/latex] into the function we get [latex]\\frac{0}{0}[\/latex], which is undefined.<\/p>\r\n

        Factoring and canceling is a good strategy.<\/p>\r\n

        [latex]\\underset{x\\to 3}{\\lim}\\dfrac{x^2-3x}{2x^2-5x-3}=\\underset{x\\to 3}{\\lim}\\dfrac{x(x-3)}{(x-3)(2x+1)}[\/latex]<\/div>\r\n

         <\/p>\r\n

        For all [latex]x\\ne 3, \\, \\frac{x^2-3x}{2x^2-5x-3}=\\frac{x}{2x+1}[\/latex]. Therefore,<\/p>\r\n

        [latex]\\underset{x\\to 3}{\\lim}\\dfrac{x(x-3)}{(x-3)(2x+1)}=\\underset{x\\to 3}{\\lim}\\dfrac{x}{2x+1}[\/latex]<\/div>\r\n

         <\/p>\r\n

        Evaluate using the limit laws:<\/p>\r\n

        [latex]\\underset{x\\to 3}{\\lim}\\dfrac{x}{2x+1}=\\dfrac{3}{7}[\/latex]
        \r\n[\/hidden-answer]<\/div>\r\n<\/section>\r\n
        \r\n

        [ohm_question hide_question_numbers=1]288277-288278[\/ohm_question]<\/p>\r\n<\/section>\r\n

        Evaluating a Limit by Multiplying by a Conjugate<\/h3>\r\n

        To tackle the limit involving a square root, we'll utilize the method of multiplying by a conjugate. This means if we have a term like [latex]\\sqrt{x}+a[\/latex], we multiply by [latex]\\sqrt{x}+-[\/latex] to rationalize and simplify the expression. This technique often resolves indeterminate forms by eliminating the radical in the denominator, allowing us to find the limit using algebraic simplification.<\/p>\r\n

        \r\n

        How To: Rationalize a Limit Involving Square Roots<\/strong><\/p>\r\n

          \r\n\t
        1. Identify the term with the square root in the limit expression.<\/li>\r\n\t
        2. Multiply the expression by the conjugate of the term with the square root, both in the numerator and denominator.<\/li>\r\n\t
        3. Expand the product to eliminate the square root.<\/li>\r\n\t
        4. Simplify the resulting expression by combining like terms and canceling where possible.<\/li>\r\n\t
        5. Apply limit laws to the simplified expression to evaluate the limit.<\/li>\r\n<\/ol>\r\n<\/section>\r\n
          \r\n

          Evaluate [latex]\\underset{x\\to -1}{\\lim}\\dfrac{\\sqrt{x+2}-1}{x+1}[\/latex]<\/p>\r\n

          [reveal-answer q=\"fs-id1170572307671\"]Show Solution[\/reveal-answer]
          \r\n[hidden-answer a=\"fs-id1170572307671\"]<\/p>\r\n

          [latex]\\frac{\\sqrt{x+2}-1}{x+1}[\/latex] has the form [latex]\\frac{0}{0}[\/latex] at [latex]\u22121[\/latex].<\/p>\r\n

          Let\u2019s begin by multiplying by [latex]\\sqrt{x+2}+1[\/latex], the conjugate of [latex]\\sqrt{x+2}-1[\/latex], on the numerator and denominator:<\/p>\r\n

          [latex]\\underset{x\\to -1}{\\lim}\\frac{\\sqrt{x+2}-1}{x+1}=\\underset{x\\to -1}{\\lim}\\frac{\\sqrt{x+2}-1}{x+1}\\cdot \\frac{\\sqrt{x+2}+1}{\\sqrt{x+2}+1}[\/latex]<\/div>\r\n

          We then multiply out the numerator. We don\u2019t multiply out the denominator because we are hoping that the [latex](x+1)[\/latex] in the denominator cancels out in the end:<\/p>\r\n

          [latex]=\\underset{x\\to -1}{\\lim}\\frac{x+1}{(x+1)(\\sqrt{x+2}+1)}[\/latex]<\/div>\r\n

          Then we cancel:<\/p>\r\n

          [latex]=\\underset{x\\to -1}{\\lim}\\frac{1}{\\sqrt{x+2}+1}[\/latex]<\/div>\r\n

          Last, we apply the limit laws:<\/p>\r\n

          [latex]\\underset{x\\to -1}{\\lim}\\frac{1}{\\sqrt{x+2}+1}=\\frac{1}{2}[\/latex]
          \r\n[\/hidden-answer]<\/div>\r\n<\/section>\r\n

          Evaluating a Limit by Simplifying Complex Fractions<\/h3>\r\n

          When tasked with solving the limit of a function involving a complex fraction, mastering simplification techniques becomes crucial. This process entails combining and reducing fractions within the limit expression, which often leads to a form where the limit laws can be applied.<\/p>\r\n

          \r\n

          How To: Simplify Complex Fractions in Limits<\/strong><\/p>\r\n

            \r\n\t
          1. Locate the LCD (Least Common Denominator) to combine rational expressions.<\/li>\r\n\t
          2. Factor numerators and denominators as needed.<\/li>\r\n\t
          3. Reduce the fraction to its simplest form.<\/li>\r\n\t
          4. With the simplified expression, apply limit laws to find the limit.<\/li>\r\n<\/ol>\r\n<\/section>\r\n
            \r\n

            Evaluate [latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}[\/latex]<\/p>\r\n

            [reveal-answer q=\"fs-id1170571681059\"]Show Solution[\/reveal-answer]
            \r\n[hidden-answer a=\"fs-id1170571681059\"]<\/p>\r\n

            [latex]\\frac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}[\/latex] has the form [latex]\\frac{0}{0}[\/latex] at [latex]1[\/latex].<\/p>\r\n

            We simplify the algebraic fraction by multiplying by [latex]\\frac{2(x+1)}{2(x+1)}[\/latex]:<\/p>\r\n

            [latex]\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1}=\\underset{x\\to 1}{\\lim}\\dfrac{\\frac{1}{x+1}-\\frac{1}{2}}{x-1} \\cdot \\dfrac{2(x+1)}{2(x+1)}[\/latex]<\/div>\r\n

            Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor [latex](x-1)[\/latex]:<\/p>\r\n

            [latex]=\\underset{x\\to 1}{\\lim}\\frac{2-(x+1)}{2(x-1)(x+1)}[\/latex]<\/div>\r\n

            Then, we simplify the numerator:<\/p>\r\n

            [latex]=\\underset{x\\to 1}{\\lim}\\frac{-x+1}{2(x-1)(x+1)}[\/latex]<\/div>\r\n

            Now we factor out [latex]\u22121[\/latex] from the numerator:<\/p>\r\n

            [latex]=\\underset{x\\to 1}{\\lim}\\frac{-(x-1)}{2(x-1)(x+1)}[\/latex]<\/div>\r\n

            Then, we cancel the common factors of [latex](x-1)[\/latex]:<\/p>\r\n

            [latex]=\\underset{x\\to 1}{\\lim}\\frac{-1}{2(x+1)}[\/latex]<\/div>\r\n

            Last, we evaluate using the limit laws:<\/p>\r\n

            [latex]\\underset{x\\to 1}{\\lim}\\frac{-1}{2(x+1)}=-\\frac{1}{4}[\/latex]
            \r\n[\/hidden-answer]<\/div>\r\n<\/section>\r\n

            The example below 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.<\/p>\r\n

            \r\n

            Evaluate [latex]\\underset{x\\to 0}{\\lim}\\left(\\dfrac{1}{x}+\\dfrac{5}{x(x-5)}\\right)[\/latex]<\/p>\r\n

            [reveal-answer q=\"fs-id1170571648205\"]Show Solution[\/reveal-answer]
            \r\n[hidden-answer a=\"fs-id1170571648205\"]<\/p>\r\n

            Both [latex]\\frac{1}{x}[\/latex] and [latex]\\frac{5}{x(x-5)}[\/latex] 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.<\/p>\r\n

            In this case, we find the limit by performing addition and then applying one of our previous strategies.<\/p>\r\n

            Observe that<\/p>\r\n

            [latex]\\begin{array}{cc} \\dfrac{1}{x}+\\dfrac{5}{x(x-5)}& =\\dfrac{x-5+5}{x(x-5)} \\\\ & =\\dfrac{x}{x(x-5)}\\end{array}[\/latex]<\/div>\r\n

            Thus,<\/p>\r\n

            [latex]\\begin{array}{cc}\\underset{x\\to 0}{\\lim}\\big(\\frac{1}{x}+\\frac{5}{x(x-5)}\\big)& =\\underset{x\\to 0}{\\lim}\\frac{x}{x(x-5)} \\\\ & =\\underset{x\\to 0}{\\lim}\\frac{1}{x-5} \\\\ & =-\\frac{1}{5} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/section>","rendered":"

            Evaluating Limits Cont.<\/h2>\n

            More Limit Evaluation Techniques<\/h3>\n

            Calculating limits is straightforward for polynomials and some rational functions using direct substitution. However, when direct substitution results in an undefined form, such as [latex]\\frac{0}{0}[\/latex], further techniques are needed.<\/p>\n

            \n

            Consider the function [latex]\\frac{x^2-1}{x-1}[\/latex].<\/p>\n

            Direct substitution for [latex]x=1[\/latex] would yield the indeterminate form [latex]\\frac{0}{0}[\/latex].<\/p>\n

            By factoring [latex]f(x)[\/latex] to its simplest form,<\/p>\n

            [latex]\\begin{array}{ccc}\\hfill f(x)& =\\dfrac{x^2-1}{x-1}\\hfill \\\\ & =\\dfrac{(x-1)(x+1)}{x-1}\\hfill \\\\ & = x+1\\hfill \\end{array}[\/latex]<\/div>\n

            we see that except for [latex]x=1,[\/latex] where the function is not defined, the limit as [latex]x[\/latex] approaches [latex]1[\/latex] is the same as the value of [latex]g(x)=x+1[\/latex] at that point, which is [latex]2[\/latex].<\/p>\n

            [latex]\\begin{array}{cc}\\hfill \\underset{x\\to 1}{\\lim}\\dfrac{x^2-1}{x-1}& =\\underset{x\\to 1}{\\lim}\\dfrac{(x-1)(x+1)}{x-1}\\hfill \\\\ & =\\underset{x\\to 1}{\\lim}(x+1)\\hfill \\\\ & =2\\hfill \\end{array}[\/latex]<\/div>\n

            The graphs of these two functions are shown in Figure 1.<\/p>\n

            \"Two
            Figure 1. The graphs of [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are identical for all [latex]x\\ne 1[\/latex]. Their limits at 1 are equal.<\/figcaption><\/figure>\n<\/section>\n

            This shows how simplification can reveal a continuous limit where direct substitution fails.<\/p>\n

            When the function assumes the form [latex]\\frac{0}{0}[\/latex] upon direct substitution, it’s an indeterminate form<\/strong>.<\/p>\n

            \n

            indeterminate form in limits<\/strong><\/h3>\n

            When direct substitution in a function yields [latex]\\frac{0}{0}[\/latex], it indicates an indeterminate form requiring further analysis to calculate the limit.<\/p>\n<\/section>\n

            \n

            How To: Solve Indeterminate [latex]\\frac{0}{0}[\/latex] Limits<\/strong><\/p>\n

              \n
            1. Verify the function has the appropriate form and cannot be evaluated immediately using the limit laws.<\/li>\n
            2. Find an equivalent function [latex]h(x)=f(x)\/g(x)[\/latex] valid for all [latex]x[\/latex] near the point of interest, except at the point itself. To do this, we may need to try one or more of the following steps:\n
                \n
              1. Factor and simplify polynomials to cancel common terms.<\/li>\n
              2. For square roots, use conjugates to rationalize.<\/li>\n
              3. Simplify complex fractions.<\/li>\n<\/ol>\n<\/li>\n
              4. Apply the limit laws to the simplified expression to find the limit.<\/li>\n<\/ol>\n<\/section>\n

                Evaluate a Limit by Factoring and Simplifying Polynomials<\/h3>\n

                A key step in resolving indeterminate limits involves factoring and simplifying polynomials to cancel out common terms. By extracting the GCF, applying special product formulas for polynomials, and using grouping techniques for trinomials, we can often simplify the expressions and resolve the indeterminate forms.<\/p>\n

                \n

                How To: Evaluate a Limit by Factoring and Simplifying Polynomials<\/strong><\/p>\n

                  \n
                1. Factor the GCF out from the polynomial terms to reduce complexity.<\/li>\n
                2. Apply factoring rules for difference of squares, perfect square trinomials, and sum\/difference of cubes to break down terms further.<\/li>\n
                3. For trinomials, utilize grouping for efficient factorization.<\/li>\n
                4. After simplification, re-evaluate the limit with the new, simplified expression.<\/li>\n<\/ol>\n<\/section>\n
                  \n

                  [latex]\\begin{array}{ll} \\text{difference of squares} & a^2 - b^2 = (a+b)(a-b) \\\\ \\text{perfect square trinomial} & a^2 + 2ab + b^2 = (a+b)^2 \\\\ \\text{sum of cubes} & a^3 + b^3 = (a+b)(a^2 - ab + b^2) \\\\ \\text{difference of cubes} & a^3 - b^3 = (a-b)(a^2 + ab + b^2) \\end{array}[\/latex]<\/p>\n<\/section>\n

                  \n

                  Evaluate [latex]\\underset{x\\to 3}{\\lim}\\dfrac{x^2-3x}{2x^2-5x-3}[\/latex]<\/p>\n