Functions Question 1
Question: If $ f(x)=|x-2| $ , then
[Roorkee 1984]
Options:
A) $ \underset{x\to 2+}{\mathop{\lim }},f(x)\ne 0 $
B) $ \underset{x\to 2-}{\mathop{\lim }},f(x)\ne 0 $
C) $ \underset{x\to 2+}{\mathop{\lim }},f(x)\ne \underset{x\to 2-}{\mathop{\lim }},f(x) $
D) $ f(x) $ is continuous at $ x=2 $
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Answer:
Correct Answer: D
Solution:
Here $ f(2)=0 $ $ \underset{x\to 2-}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }}f(2-h)=\underset{h\to 0}{\mathop{\lim }},|2-h-2|=0 $ $ \underset{x\to 2-}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }}f(2-h)=\underset{h\to 0}{\mathop{\lim }},|2-h-2|=0 $ Hence it is continuous at $ x=2 $ .
BETA
