If-f-x-xtan-1-1-x-x-0-0-x-0-show-that-f-is-countinous-but-not-differentiable-at-x-0-

Question Number 11608 by agni5 last updated on 29/Mar/17

If f (x) = {xtan}^{- 1} (\frac{1}{x}), x \neq 0

= 0, x = 0

show that f is countinous but not differentiable

at x = 0 .

Answered by mrW1 last updated on 30/Mar/17

f (0) = 0

\lim_{x \to 0} f (x) = \lim_{x \to 0} x \tan^{- 1} (\frac{1}{x}) = (\lim_{x \to 0} x) \times (\lim_{x \to 0} \tan^{- 1} \frac{1}{x})

= 0 \times (\pm \frac{π}{2}) = 0

s i n c e f (0) = \lim_{x \to 0} f (x)

\Rightarrow f (x) i s c o n t i n o u s a t x = 0 .

f^{'} (x) = \tan^{- 1} (\frac{1}{x}) + x (\frac{1}{1 + \frac{1}{x^{2}}}) (- \frac{1}{x^{2}}) = \tan^{- 1} (\frac{1}{x}) - \frac{x}{1 + x^{2}}

\lim_{x \to - 0} f^{'} (x) = - \frac{π}{2} - 0 = - \frac{π}{2}

\lim_{x \to + 0} f^{'} (x) = \frac{π}{2} - 0 = \frac{π}{2}

s i n c e \lim_{x \to - 0} f^{'} (x) \neq \lim_{x \to + 0} f^{'} (x)

\Rightarrow f (x) i s n o t d i f f e r e n t i a b l e a t x = 0 .

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