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

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:

{\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.