Calculus

Limits and Continuity of Functions

Limits and Continuity Logo

Use of Infinitesimals

  • The function \(\alpha \left( x \right)\) is called infinitely small or an infinitesimal as \(x \to a\) if

    \[\lim\limits_{x \to a} \alpha \left( x \right) = 0.\]

    Let \(\alpha \left( x \right)\) and \(\beta \left( x \right)\) be two infinitely small functions as \(x \to a.\)

    • If \(\lim\limits_{x \to a} {\large\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}\normalsize} = 0,\) then we say that the function \(\alpha \left( x \right)\) is an infinitesimal of higher order than \(\beta \left( x \right);\)
    • If \(\lim\limits_{x \to a} {\large\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}\normalsize} = A \ne 0,\) then the functions \(\alpha \left( x \right)\) and \(\beta \left( x \right)\) are called infinitesimals of the same order;
    • If \(\lim\limits_{x \to a} {\large\frac{{\alpha \left( x \right)}}{{{\beta ^n}\left( x \right)}}\normalsize} = A \ne 0,\) then the function \(\alpha \left( x \right)\) is called an infinitesimal of order \(n\) compared with the function \(\beta \left( x \right);\)
    • If \(\lim\limits_{x \to a} {\large\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}\normalsize} = 1,\) then the functions \(\alpha \left( x \right)\) and \(\beta \left( x \right)\) are said to be equivalent as \(x \to a.\)

    In particular, the following functions are equivalent:

    equivalent infinitesimals

    When calculating the limit of a ratio of two infinitesimals, we can replace the terms of the ratio by their equivalent values.


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Find the limit \(\lim\limits_{x \to 0} {\large\frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}}\normalsize}.\)

    Example 2

    Find the limit \(\lim\limits_{x \to 0} {\large\frac{{\sqrt[3]{{1 + x}} – 1}}{x}\normalsize}.\)

    Example 3

    Find the limit \(\lim\limits_{t \to 0} {\large\frac{{1 – \cos \left( {1 – \cos t} \right)}}{{{{\sin }^2}{t^2}}}\normalsize}.\)

    Example 4

    Calculate the limit \(\lim\limits_{x \to 0} {\large\frac{{\sqrt {1 + 2x + 3{x^2}} – 1}}{x}\normalsize}.\)

    Example 5

    Calculate the limit \(\lim\limits_{x \to e} {\large\frac{{\ln \left( {\ln x} \right)}}{{x – e}}\normalsize}.\)

    Example 6

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

    Example 7

    Calculate the limit \(\lim\limits_{x \to 2} {\large\frac{{{{\log }_2}x – 1}}{{x – 2}}\normalsize}.\)

    Example 8

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

    Example 9

    Calculate the limit \(\lim\limits_{x \to 0} {\large\frac{{\ln \cos x}}{{\sqrt[3]{{1 + {x^2}}} – 1}}\normalsize}.\)

    Example 10

    Calculate the limit \(\lim\limits_{t \to a} {\left( {\large\frac{{\sin t}}{{\sin a}}\normalsize} \right)^{\large\frac{1}{{t – a}}\normalsize}}.\)

    Example 1.

    Find the limit \(\lim\limits_{x \to 0} {\large\frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}}\normalsize}.\)

    Solution.

    We use the formulas:

    \[\ln \left( {1 + \alpha } \right) \sim \alpha ,\;\;\;\sin \alpha \sim \alpha .\]

    Then

    \[{\lim\limits_{x \to 0} \frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}} = \lim\limits_{x \to 0} \frac{{4x}}{{3x}} }={ \frac{4}{3}.}\]

    Example 2.

    Find the limit \(\lim\limits_{x \to 0} {\large\frac{{\sqrt[3]{{1 + x}} – 1}}{x}\normalsize}.\)

    Solution.

    As \(\sqrt[\large 3\normalsize]{{1 + x}} \sim 1 + {\large\frac{x}{3}\normalsize},\) the limit can be written as

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

    Page 1
    Problems 1-2
    Page 2
    Problems 3-10