Question

Discuss the continuity and differentiability of 

\[f\left( x \right) = \begin{cases}\left( x - c \right) \cos \left( \frac{1}{x - c} \right), & x \neq c \\ 0 , & x = c\end{cases}\]

Answer 1

Given:  

\[f\left( x \right) = \begin{cases}\left( x - c \right) \cos \left( \frac{1}{x - c} \right), & x \neq c \\ 0 , & x = c\end{cases}\]

Continuity:
(LHL at x = c) 

\[\lim_{x \to c^-} f(x) \]
\[ = \lim_{h \to 0} f(c - h) \]
\[ = \lim_{h \to 0} (c - h - c) \cos\left( \frac{1}{c - h - c} \right)\]
\[ = \lim_{h \to 0} - h \cos\left( \frac{1}{h} \right) \]
\[\text { Since , cos } \left( \frac{1}{h} \right) \text{is a bounded function and 0 × times bounded function is} 0\]

(RHL at x = c)  

\[\lim_{x \to c^+} f(x) \]
\[ = \lim_{h \to 0} f(c + h) \]
\[ = \lim_{h \to 0} (c + h - c) \cos\left( \frac{1}{c + h - c} \right)\]
\[ = \lim_{h \to 0} h \cos\left( \frac{1}{h} \right) \]
\[\text { Since} , \cos\left( \frac{1}{h} \right) \text{is a bounded function and 0 times bounded function is} 0\]

and 
Differentiability at x = c

(LHD at x = c)

\[\lim_{x \to c^-} \frac{f(x) - f(c)}{x - c} \]
\[ = \lim_{h \to 0} \frac{f(c - h) - f(c)}{c - h - c} \]
\[ = \lim_{h \to 0} \frac{- h \cos\left( \frac{1}{- h} \right) - 0}{- h} \left[ \because 0 . \cos \left( \frac{1}{c - c} \right) = 0, as \cos\text {  function is bounded function }. \right]\]
\[ = \lim_{h \to 0} \cos\left( \frac{1}{h} \right)\]
\[ = \text { A number which oscillates between - 1 and 1 }\]
\[ \therefore \text { LHD } \hspace{0.167em} (x = c) \text { does not exist } . \]
\[\text{Similarly , we can show that RHD(x = c) does not exist} . \]
\[\text{Hence , f(x) is not differentiable at x} = c\]