Question

Show that f(x) = |x − 2| is continuous but not differentiable at x = 2. 

Answer 1

Given:  

\[f(x) = |x - 2| = \] `{(x-2,x≥,2),(-x+2,x<,2) :}`

 

Continuity at x=2: We have,
(LHL at x = 2) 

\[{= \lim}_{x \to 2^-} f(x) \]
\[ = \lim_{h \to 0} f(2 - h) \]
\[ = \lim_{h \to 0} ( - 2 + h) + 2\]
\[ = 0\]

(RHL at x = 2) 

\[{= \lim}_{x \to 2^+} f(x) \]
\[ = \lim_{h \to 0} f(2 + h) \]
\[ = \lim_{h \to 0} 2 + h - 2 \]
\[ = 0\]
and 

\[f(2) = 0\]

\[\lim_{x \to 2^-} f(x)\]

\[\lim_{x \to 2^+} f(x)\]

\[f(2)\]

Hence, 

\[f(x)\]  is continuous at 

\[x = 2\]

Differentiability at x = 2: We have,

(LHD at x = 2)

\[{=lim}_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} \]
\[ = \lim_{x \to 2} \frac{( - x + 2) - 0}{x - 2} \]
\[ = \lim_{x \to 2} \frac{- (x - 2)}{x - 2} \]
\[ = \lim_{x \to 2} ( - 1) \]
\[ = - 1\]

(RHD at x=2)
= 

\[= \lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} \]
\[ = \lim_{x \to 2} \frac{(x - 2) - 0}{x - 2} \]
\[ = \lim_{x \to 2} 1 \]
\[ = 1\]

Thus, 

\[\lim_{x \to 2^-} f(x)\]

\[\lim_{x \to 2^+} f(x)\]

Hence,  

\[f(x)\] is not differentiable at x=2 .