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Calculus Level 3

f ( x ) = { sin ⁡ x x , x < 0 x + 1 , x ≥ 0 \large f(x) = \begin{cases} \dfrac{\sin x}{x}, & x < 0 \\ x + 1, & x \ge 0 \end{cases}

Find point of discontinuity of the function above (if any).

Always continuous At x = 1 x = 1 At x = − 1 x = -1 At x = 0 x = 0

Nikhil Raj by Nikhil Raj , 19, India

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1 solution

Nikhil Raj
Jun 1, 2017

f ( x ) = { = s i n x x when x < 0 = x + 1 when x ≥ 0 So, only doubtful point for continuity of f ( x ) is x = 0. S o , f ( 0 ) = 0 + 1 = 1 Left hand limit = lim ⁡ h → 0 f ( 0 − h ) = lim ⁡ h → 0 s i n ( 0 − h ) ( 0 − h ) = 1 Right hand limit = lim ⁡ h → 0 f ( 0 + h ) = lim ⁡ h → 0 s i n ( 0 + h ) ( 0 + h ) = 1 S i n c e , f ( 0 ) = L . H . L = R . H . L . , So, Given function is continuous everywhere \large f(x) = \begin{cases} = \dfrac{sin x}{x} {\text{ when }} x < 0 \\ = x + 1 {\text{ when }} x \ge 0 \end{cases} \\ {\text{So, only doubtful point for continuity of }} f(x) {\text{ is }} x = 0. \\ So, f(0) = 0 + 1 = 1 \\ {\text{Left hand limit}} = {\displaystyle{\lim_{h \rightarrow 0}}} f(0 - h) = {\displaystyle{\lim_{h \rightarrow 0}}} \dfrac{sin (0 - h)}{(0 - h)} = 1 \\ {\text{Right hand limit}} = {\displaystyle{\lim_{h \rightarrow 0}}} f(0 + h) = {\displaystyle{\lim_{h \rightarrow 0}}} \dfrac{sin (0 + h)}{(0 + h)} = 1 \\ Since, f(0) = L.H.L =R.H.L. , {\text{So, Given function is}} \color{#E81990}{\boxed{\text{continuous everywhere}}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

That's gotcha kind of an answer. Two options are correct. But if you give one of the correct answers, you are wrong. Not fun.

Marta Reece - 4 years ago

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I agree. The trickery in the options does not add to the value of the question. In fact, if "two options are correct" were the correct answer then this implies that there are multiple correct options, which means that the answer is not unique and hence the question is flawed.

Brian Charlesworth - 4 years ago

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Even more problematic, the answer is self-referential and in doing so essentially suffers from the liar's paradox.

Brian Moehring - 4 years ago

Seconded. There should be only one correct option.

Jon Haussmann - 4 years ago

First read the options, then attempt the answer.

Nikhil Raj - 4 years ago

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