Calculus

Limits and Continuity of Functions

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L’Hopital’s Rule

  • L’Hopital’s Rule provides a method for evaluating indeterminate forms of type \(\large\frac{0}{0}\normalsize\) or \(\large\frac{\infty}{\infty}\normalsize.\)

    Let \(a\) be either a finite number or infinity.

    • If \(\lim\limits_{x \to a} f\left( x \right) = 0\) and \(\lim\limits_{x \to a} g\left( x \right) = 0,\) then \(\lim\limits_{x \to a} {\large\frac{{f\left( x \right)}}{{g\left( x \right)}}\normalsize} = \lim\limits_{x \to a} {\large\frac{{f’\left( x \right)}}{{g’\left( x \right)}}\normalsize};\)
    • If \(\lim\limits_{x \to a} f\left( x \right) = \infty\) and \(\lim\limits_{x \to a} g\left( x \right) = \infty,\) then similarly \(\lim\limits_{x \to a} {\large\frac{{f\left( x \right)}}{{g\left( x \right)}}\normalsize} = \lim\limits_{x \to a} {\large\frac{{f’\left( x \right)}}{{g’\left( x \right)}}\normalsize}.\)

    This rule appeared in \(1696\) (!) in the first book on differential calculus published by French mathematician Guillaume de l’Hopital \(\left(1661- 1704\right).\)

    Guillaume de l'Hopital
    Fig.1 Guillaume de l’Hopital (1661-1704)

    We can apply L’Hopital’s rule to indeterminate forms of type \(0 \cdot \infty,\) \(\infty – \infty,\) \(0^0,\) \(1^{\infty},\) \(\infty^0\) as well. The first two indeterminate forms \(0 \cdot \infty\) and \(\infty – \infty\) can be reduced to the forms \(\large\frac{0}{0}\normalsize\) or \(\large\frac{\infty}{\infty}\normalsize\) using algebraic transformations. The indeterminate forms \(0^0,\) \(1^{\infty}\) and \(\infty^0\) can be reduced to the form \(0 \cdot \infty\) with the help of identity

    \[f{\left( x \right)^{g\left( x \right)}} = {e^{g\left( x \right)\ln f\left( x \right)}}.\]

    L’Hopital’s rule also holds for one-sided limits.


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Find the limit \(\lim\limits_{x \to 0} {\large\frac{{{a^x} – 1}}{x}\normalsize}.\)

    Example 2

    Find the limit \(\lim\limits_{x \to 2} {\large\frac{{\sqrt {7 + x} – 3}}{{x – 2}}\normalsize}.\)

    Example 3

    Calculate the limit \(\lim\limits_{x \to 2} \left( {{\large\frac{4}{{{x^2} – 4}}\normalsize} – {\large\frac{1}{{x – 2}}\normalsize}} \right).\)

    Example 4

    Find the limit \(\lim\limits_{x \to \infty } {\large\frac{{{x^2}}}{{{2^x}}}\normalsize}.\)

    Example 5

    Calculate the limit \(\lim\limits_{t \to \infty } \sqrt[\large t\normalsize]{{{t^3}}}.\)

    Example 6

    Find the limit \(\lim\limits_{x \to 1} {x^{\large\frac{1}{{1 – x}}\normalsize}}.\)

    Example 7

    Calculate the limit \(\lim\limits_{x \to 0 + } {x^x}.\)

    Example 8

    Calculate the limit \(\lim\limits_{x \to 0 + } {\left( {\arcsin x} \right)^{2x}}.\)

    Example 9

    Find the limit \(\lim\limits_{x \to \pi /2} {\left( {\sin x} \right)^{\tan x}}.\)

    Example 10

    Calculate the limit \(\lim\limits_{x \to + \infty } {\large\frac{{{x^n}}}{{{e^x}}}\normalsize}.\)

    Example 11

    Find the limit \(\lim\limits_{x \to \pi /2} {\large\frac{{\tan x}}{{\tan 3x}}\normalsize}.\)

    Example 12

    Calculate the limit \(\lim\limits_{x \to \pi /2} {\left( {\tan x} \right)^{\cos x}}.\)

    Example 1.

    Find the limit \(\lim\limits_{x \to 0} {\large\frac{{{a^x} – 1}}{x}\normalsize}.\)

    Solution.

    Differentiating the numerator and denominator separately, we get

    \[
    {\lim\limits_{x \to 0} \frac{{{a^x} – 1}}{x} = \left[ {\frac{0}{0}} \right] }
    = {\lim\limits_{x \to 0} \frac{{{{\left( {{a^x} – 1} \right)}^\prime }}}{{{{\left( x \right)}^\prime }}} }
    = {\lim\limits_{x \to 0} \frac{{{a^x}\ln a}}{1} = \ln a\lim\limits_{x \to 0} {a^x} }
    = {\ln a \cdot 1 = \ln a.}
    \]

    Example 2.

    Find the limit \(\lim\limits_{x \to 2} {\large\frac{{\sqrt {7 + x} – 3}}{{x – 2}}\normalsize}.\)

    Solution.

    Because direct substitution leads to an indeterminate form \(\large\frac{0}{0}\normalsize,\) we can use L’Hopital’s rule:

    \[
    {\lim\limits_{x \to 2} \frac{{\sqrt {7 + x} – 3}}{{x – 2}} = \left[ {\frac{0}{0}} \right] }
    = {\lim\limits_{x \to 2} \frac{{{{\left( {\sqrt {7 + x} – 3} \right)}^\prime }}}{{{{\left( {x – 2} \right)}^\prime }}} }
    = {\lim\limits_{x \to 2} \frac{{\large\frac{1}{{2\sqrt {7 + x} }}\normalsize}}{1} }
    = {\frac{1}{2}\lim\limits_{x \to 2} \frac{1}{{\sqrt {7 + x} }} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}}
    \]

    Example 3.

    Calculate the limit \(\lim\limits_{x \to 2} \left( {{\large\frac{4}{{{x^2} – 4}}\normalsize} – {\large\frac{1}{{x – 2}}\normalsize}} \right).\)

    Solution.

    Here we deal with an inderminate form of type \(\infty – \infty.\) After simple transformations, we have

    \[
    {\lim\limits_{x \to 2} \left( {\frac{4}{{{x^2} – 4}} – \frac{1}{{x – 2}}} \right) }
    = {\lim\limits_{x \to 2} \frac{{4 – \left( {x + 2} \right)}}{{{x^2} – 4}} }
    = {\lim\limits_{x \to 2} \frac{{2 – x}}{{{x^2} – 4}} = \left[ {\frac{0}{0}} \right] }
    = {\lim\limits_{x \to 2} \frac{{{{\left( {2 – x} \right)}^\prime }}}{{{{\left( {{x^2} – 4} \right)}^\prime }}} }
    = {\lim\limits_{x \to 2} \left( {\frac{{ – 1}}{{2x}}} \right) = – \frac{1}{4}.}
    \]

    Page 1
    Problems 1-3
    Page 2
    Problems 4-12