Calculus

Limits and Continuity of Functions

Limits and Continuity Logo

Discontinuous Functions – Page 2

  • Example 2.

    Show that the function \(f\left( x \right) = {\large\frac{{\sin x}}{x}\normalsize}\) has a removable discontinuity at \(x = 0.\)

    Solution.

    Obviously, the function is not defined at \(x = 0\). As \(\sin x\) is continuous at every \(x\), then the initial function \(f\left( x \right) = {\large\frac{{\sin x}}{x}\normalsize}\) is also continuous for all \(x\) except the point \(x = 0\).

    Since \(\lim\limits_{x \to 0} {\large\frac{{\sin x}}{x}\normalsize} = 1,\) the function has a removable discontinuity at this point. We can construct the new function

    \[
    {f_1}\left( x \right) =
    \begin{cases}
    \large\frac {\sin x}{x}\normalsize, & x \ne 0 \
    1, &x = 0
    \end{cases}
    ,\]

    which is continuous at every real \(x.\)

    Example 3.

    Find the points of discontinuity of the function \(f\left( x \right) = \begin{cases} 1 – {x^2}, & x \lt 0 \\ x +2, &x \ge 0 \end{cases} \) if they exist.

    Solution.

    The function exists for all \(x,\) however it is defined by two different functions and, therefore, is not elementary. We investigate “behavior” of the function near to the point \(x = 0\) where its analytic expression changes.

    Calculate one-sided limits at \(x = 0.\)

    \[
    {\lim\limits_{x \to 0 – 0} f\left( x \right) = \lim\limits_{x \to 0 – 0} \left( {1 – {x^2}} \right) = 1,\;\;\;}\kern-0.3pt
    {\lim\limits_{x \to 0 + 0} f\left( x \right) = \lim\limits_{x \to 0 – 0} \left( {x + 2} \right) = 2.}
    \]

    Thus, the function has a discontinuity of the first kind at \(x = 0.\) The finite jump at this point is

    \[
    {\Delta y = \lim\limits_{x \to 0 + 0} f\left( x \right) – \lim\limits_{x \to 0 – 0} f\left( x \right) }\kern0pt{= 2 – 1 = 1.}
    \]

    The function is continuous for all other \(x,\) because both the functions defined from the left and from the right of the point \(x = 0\) are elementary functions without any discontinuities.

    Example 4.

    Find the points of discontinuity of the function \(f\left( x \right) = \arctan {\large\frac{1}{x}\normalsize}\) if they exist.

    Solution.

    This elementary function is defined for all \(x\) except \(x = 0\), where it has a discontinuity. Find one-sided limits:

    \[
    {\lim\limits_{x \to 0 – 0} \arctan \frac{1}{x} = \arctan \left( { – \infty } \right) = – \frac{\pi }{2},\;\;\;}\kern-0.3pt
    {\lim\limits_{x \to 0 + 0} \arctan \frac{1}{x} = \arctan \left( { + \infty } \right) = \frac{\pi }{2}.}
    \]

    As seen, the function has a discontinuity point of the first kind at \(x = 0\) (Figure \(5\)). The jump of the function is equal to \(\pi.\)

    function inverse tangent of 1/x
    Figure 5.

    Example 5.

    Find the points of discontinuity of the function \(f\left( x \right) = {\large\frac{{\left| {2x + 5} \right|}}{{2x + 5}}\normalsize}\) if they exist.

    Solution.

    The function is defined and continuous for all \(x\) except \(x = – {\large\frac{5}{2}\normalsize}\), where it has a discontinuity. Investigate this discontinuity point:

    \[
    {\lim\limits_{x \to – \frac{5}{2} – 0} \frac{{\left| {2x + 5} \right|}}{{2x + 5}} }
    = {\lim\limits_{x \to – \frac{5}{2} – 0} \frac{{ – \left( {2x + 5} \right)}}{{2x + 5}} = – 1,\;\;\;}\kern-0.3pt
    {\text{if}\;\;x \lt – \frac{5}{2};}
    \]

    \[
    {\lim\limits_{x \to – \frac{5}{2} + 0} \frac{{\left| {2x + 5} \right|}}{{2x + 5}} }
    = {\lim\limits_{x \to – \frac{5}{2} + 0} \frac{{ \left( {2x + 5} \right)}}{{2x + 5}} = 1,\;\;\;}\kern-0.3pt
    {\text{if}\;\;x \ge – \frac{5}{2}.}
    \]

    As the values of the one-sided limits are finite, then there’s a discontinuity of the first kind at the point \(x = – {\large\frac{5}{2}\normalsize}.\) The graph of the function is sketched in Figure \(6.\)

    jump discontinuity
    Figure 6.
    Page 1
    Problem 1
    Page 2
    Problems 2-5