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Question
Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\]
Answer in Brief
Solution
\[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2}\]
\[LHL: \]
\[ \lim_{x \to 2^-} \left( \frac{\left| x - 2 \right|}{x - 2} \right)\]
\[Let x = 2 - h\]
\[\text{ If } x \to 2, \text{ then } h \to 0 . \]
\[ = \lim_{h \to 0} \frac{\left| 2 - h - 2 \right|}{2 - h - 2} = - 1\]
\[RHL: \]
\[ \lim_{x \to 2^+} \left( \frac{\left| x - 2 \right|}{x - 2} \right)\]
\[\text{ Let } x = 2 + h\]
\[\text{ If } x \to 2, \text{ then } h \to 0 . \]
\[ \lim_{h \to 0} \frac{\left| 2 + h - 2 \right|}{2 + h - 2} = 1\]
\[LHL \neq RHL\]
\[\text{ Thus }, \lim_{x \to 2} \left( \frac{\left| x - 2 \right|}{x - 2} \right) \text{ does not exist } .\]
\[LHL: \]
\[ \lim_{x \to 2^-} \left( \frac{\left| x - 2 \right|}{x - 2} \right)\]
\[Let x = 2 - h\]
\[\text{ If } x \to 2, \text{ then } h \to 0 . \]
\[ = \lim_{h \to 0} \frac{\left| 2 - h - 2 \right|}{2 - h - 2} = - 1\]
\[RHL: \]
\[ \lim_{x \to 2^+} \left( \frac{\left| x - 2 \right|}{x - 2} \right)\]
\[\text{ Let } x = 2 + h\]
\[\text{ If } x \to 2, \text{ then } h \to 0 . \]
\[ \lim_{h \to 0} \frac{\left| 2 + h - 2 \right|}{2 + h - 2} = 1\]
\[LHL \neq RHL\]
\[\text{ Thus }, \lim_{x \to 2} \left( \frac{\left| x - 2 \right|}{x - 2} \right) \text{ does not exist } .\]
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