Question

Choose the correct option(s) for the following question given below:
Function whose jump (non-negative difference of \[LHL\]and \[RHL\]) of discontinuity is greater than or equal to one, is are-
A. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\left( {{e^{\dfrac{1}{x}}} + 1} \right)}}{{\left( {{e^{\dfrac{1}{x}}} - 1} \right)}},\,\,\,\,\,x < 0} \\
  {\dfrac{{\left( {1 - \cos x} \right)}}{x},\,\,\,\,\,x < 0}
\end{array}} \right.\]
B. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\left( {{x^{\dfrac{1}{3}}} + 1} \right)}}{{\left( {{x^{\dfrac{1}{2}}} - 1} \right)}},\,\,\,\,\,x > 1} \\
  {\dfrac{{\ln x}}{{\left( {x - 1} \right)}},\,\,\,\,\,\dfrac{1}{2} < x < 1}
\end{array}} \right.\]
C. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{{{\sin }^{ - 1}}2x}}{{{{\tan }^{ - 1}}3x}},\,\,\,\,\,x \in \left( {0,\left. {\dfrac{1}{2}} \right]} \right.} \\
  {\dfrac{{\left| {\sin x} \right|}}{x},\,\,\,\,\,x < 0}
\end{array}} \right.\]
D. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {{{\log }_3}\left( {x + 2} \right),\,\,\,\,\,x > 2} \\
  {{{\log }_{\dfrac{1}{2}}}\left( {{x^2} + 5} \right),\,\,\,\,\,x < 2}
\end{array}} \right.\]

Answer 1

Hint: We are going to solve this question by trial and error method. We have to take each of the given options into checking. First, we check the left-hand limit and then the right-hand limit. Then we find the difference between both the limits, which is the jump of the discontinuity. According to the jump, we conclude the final asked solution.

Complete answer:
Given, four options for us to check the asked condition. Let us check by each option.
Option A:
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\left( {{e^{\dfrac{1}{x}}} + 1} \right)}}{{\left( {{e^{\dfrac{1}{x}}} - 1} \right)}},\,\,\,\,\,x < 0} \\
  {\dfrac{{\left( {1 - \cos x} \right)}}{x},\,\,\,\,\,x < 0}
\end{array}} \right.\]
\[LHL = \mathop {\lim }\limits_{x \to 0} \]
\[f(x) = \mathop {\lim }\limits_{x \to 0} f\left( {0 - h} \right)\]
Substituting the limit values:
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {{e^{\dfrac{{ - 1}}{h}}} + 1} \right)}}{{\left( {{e^{\dfrac{{ - 1}}{h}}} - 1} \right)}}\]
\[ \Rightarrow \dfrac{{0 + 1}}{{0 - 1}} = - 1\]
Now taking the\[RHL\];
\[RHL = \mathop {\lim }\limits_{x \to {0^ + }} \]
\[ \Rightarrow f(x) = \mathop {\lim }\limits_{_{h \to 0}} f\left( {h + 0} \right)\]
Substituting the limits;
\[\mathop {\lim }\limits_{_{h \to 0}} \dfrac{{\left( {1 - \cos \,h} \right)}}{h}\]
Substituting another value for the above formula, we get;
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{2{{\sin }^2}\dfrac{h}{2}}}{h}\]
Simplifying we get,
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{h}{4}\dfrac{{2{{\sin }^2}\dfrac{h}{2}}}{{\dfrac{h}{2}}}\]
Substituting limit value,
\[ \Rightarrow 0 \times 1 = 0\]
\[RHL = 0\]
The difference between \[LHL\] and \[RHL\]\[ = 0 - ( - 1)\]\[ = 1\]
Therefore, the jump of the discontinuity is \[1\].

Option B:
\[\begin{gathered}
  f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\left( {{x^{\dfrac{1}{3}}} - 1} \right)}}{{\left( {{x^{\dfrac{1}{2}}} - 1} \right)}},\,\,\,\,\,x > 1} \\
  {\dfrac{{\ln x}}{{\left( {x - 1} \right)}},\,\,\,\,\,\dfrac{1}{2} < x < 1}
\end{array}} \right. \\
    \\
\end{gathered} \]
\[RHL = \mathop {\lim }\limits_{x \to {1^ + }} \]
\[f(x) = \mathop {\lim }\limits_{h \to 0} f(1 + h)\]
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{{{\left( {1 + h} \right)}^{\dfrac{1}{3}}} - 1}}{{{{\left( {1 + h} \right)}^{\dfrac{1}{2}}} - 1}}\]
Expanding the above term, we get;
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{\left( {1 + \dfrac{h}{3} - \dfrac{{{h^2}}}{9} + ...} \right) - 1}}{{\left( {1 + \dfrac{h}{2} - \dfrac{{{h^2}}}{8} + ...} \right) - 1}}\]
Taking the common terms out, we get;
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{h\left( {\dfrac{1}{3} - \dfrac{h}{9} + ...} \right)}}{{h\left( {\dfrac{1}{2} - \dfrac{h}{8} + ...} \right)}}\]
Now cancelling and approximating the above equation, we get;
\[ \Rightarrow \dfrac{2}{3}\]
\[RHL = \dfrac{2}{3}\]
Now taking the \[LHL\];
\[LHL = \mathop {\lim }\limits_{x \to {0^ + }} \]
\[f(x) = \mathop {\lim }\limits_{_{h \to 0}} f\left( {h + 0} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{\ln \left( {1 - h} \right)}}{{1 - h - 1}}\]
Simplifying we get,
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{\ln \left( {1 + \left( { - h} \right)} \right)}}{{ - h}}\]
\[\because \mathop {\lim }\limits_{_{x \to 0}} \dfrac{{\ln (1 + x)}}{x} = 1\],
Applying this value for the above equation, we get;
\[LHL = 1\]
The difference between \[LHL\] and \[RHL\] \[ = 1 - \dfrac{2}{3} = \dfrac{1}{3}\]
The answer is less than one. Hence the jump of discontinuity is not one, nor greater than one.

Option C:
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{{{\sin }^{ - 1}}2x}}{{{{\tan }^{ - 1}}3x}},\,\,\,\,\,x \in \left( {0,\left. {\dfrac{1}{2}} \right]} \right.} \\
  {\dfrac{{\left| {\sin x} \right|}}{x},\,\,\,\,\,x < 0}
\end{array}} \right.\]
\[LHL = \mathop {\lim }\limits_{_{x \to {0^ - }}} \]
\[f(x) = \mathop {\lim }\limits_{_{h \to 0}} f\left( {0 - h} \right)\]
Substituting the limit values:
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{\sin \left| {0 - h} \right|}}{{ - h}}\]
Simplifying we get
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{\left| { - \sin \,h} \right|}}{{ - h}}\]
Taking the numerator out of the modulus, we get;
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \,h}}{{ - h}}\]
Now, applying the limits, we get;
\[ \Rightarrow - 1\]
\[LHL = - 1\]
Taking the \[RHL\]
\[RHL = \mathop {\lim }\limits_{_{x \to {0^ + }}} \]
\[f(x) = \mathop {\lim }\limits_{_{h \to 0}} f(0 + h)\]
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{{{\sin }^{ - 1}}2h}}{{{{\tan }^{ - 1}}3h}}\]
Multiplying the terms in the numerator and denominator;
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} \dfrac{{2h\dfrac{{{{\sin }^{ - 1}}2h}}{{2h}}}}{{3h\dfrac{{{{\tan }^{ - 1}}3h}}{{3h}}}}\]
Cancelling out the common terms and applying the limits, we get;
\[RHL = \dfrac{2}{3}\]
The difference between \[LHL\] and \[RHL\] \[ = \dfrac{2}{3} - 1\] \[ = \dfrac{5}{3} > 1\]
Therefore, the jump of discontinuity for the function is greater than one.

Option D:
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {{{\log }_3}\left( {x + 2} \right),\,\,\,\,\,x > 2} \\
  {{{\log }_{\dfrac{1}{2}}}\left( {{x^2} + 5} \right),\,\,\,\,\,x < 2}
\end{array}} \right.\]
\[RHL = \mathop {\lim }\limits_{x \to {2^ + }} \]
\[f(x) = \mathop {\lim }\limits_{_{h \to 0}} f(2 + h)\]
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} {\log _3}\left( {h + 4} \right)\]
Applying the limits, we get;
\[{\log _3}4 = 1.26\]
\[RHL = 1.26\]
Now, taking \[LHL\]
\[LHL = \mathop {\lim }\limits_{_{x \to {2^ - }}} \]
\[f(x) = \mathop {\lim }\limits_{_{h \to 0}} f(2 - h)\]
\[ \Rightarrow \mathop {\lim }\limits_{_{h \to 0}} {\log _{\dfrac{1}{2}}}\left( {{h^2} - 4h + 9} \right)\]
Simplifying the above term, we get;
\[ \Rightarrow {\log _{\dfrac{1}{2}}}9 = - 0.653\]
The difference between \[LHL\] and \[RHL\] \[ = 1.26 - ( - 0.653) > 1\]
Hence, the jump of discontinuity for the function is greater than one.

Hence, the correct answers are option (A), (C) and (D).

Note: We have to remember two things here, one is jump discontinuity and another one is a trial and error method.
A jump discontinuity is when the two-sided limit doesn’t exist because the one-sided limits aren’t equal. Asymptotic or infinite discontinuity is when the two-sided limit doesn’t exist because it’s unbounded.
Trial and error refers to the process of verifying that a certain choice is right (or wrong). We simply substitute that choice into the problem and check.