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Show that F ( X ) = { | X − a | X − a , W H E N X ≠ a 1 , W H E N X = a is Discontinuous at X = A. - Mathematics

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प्रश्न

Show that 

\[f\left( x \right) = \begin{cases}\frac{\left| x - a \right|}{x - a}, when & x \neq a \\ 1 , when & x = a\end{cases}\] is discontinuous at x = a.

उत्तर

The given function can be rewritten as: 

\[f\left( x \right) = \begin{cases}\frac{x - a}{x - a}, when x > a \\ \frac{a - x}{x - a}, when x < a \\ 1, when x = a\end{cases}\]

\[\Rightarrow\] \[f\left( x \right) = \begin{cases}1, when x > a \\ - 1, when x < a \\ 1, when x = a\end{cases}\]
\[\Rightarrow\] \[f\left( x \right) = \binom{1, when x \geq a}{ - 1, when x < a}\]
We observe
(LHL at x = a) = \[\lim_{x \to a^-} f\left( x \right) = \lim_{h \to 0} f\left( a - h \right)\]
\[= \lim_{h \to 0} \left( - 1 \right) = - 1\]
(RHL at x = a) = 
\[\lim_{x \to a^+} f\left( x \right) = \lim_{h \to 0} f\left( a + h \right)\]
\[\lim_{h \to 0} \left( 1 \right) = 1\]
\[\lim_{x \to a^-} f\left( x \right) \neq \lim_{x \to a^+} f\left( x \right)\]

Thus, f(x) is discontinuous at x = a.

 
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अध्याय 9: Continuity - Exercise 9.1 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 9 Continuity
Exercise 9.1 | Q 9 | पृष्ठ १७

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