why-sinx-is-not-differentiable-at-x-npi-

Question Number 134519 by sachin1221 last updated on 04/Mar/21

w h y s i n x i s n o t d i f f e r e n t i a b l e a t x = n π

Commented by JDamian last updated on 04/Mar/21

seriously?

Answered by physicstutes last updated on 04/Mar/21

\lim_{x \to 0} (\frac{\sin x - 0}{x - 0}) = \lim_{x \to 0} \frac{\sin x}{x} = 1

\sin x is also continuous and so it is differentiable at

x = 0 . what do you mean by your question ?

Commented by sachin1221 last updated on 04/Mar/21

\lim_{x \to 0} (\frac{\sin x - 0}{x - 0}) = \lim_{x \to 0} \frac{\sin x}{x} = 1

\sin x is also continuous and so it is differentiable at

x = 0 . what do you mean by your question ?

b u t w h e n u s e e g r a p h i c a l m e t h o d a t x = 0 u w i l l g e t i n f i n i t e t a n g e n t a t x = 0 s o t h e r e i s n o d i f f e r e n t i a b l e a t x = 0

Commented by mr W last updated on 05/Mar/21

c a n y o u s h o w u s h o w y o u r g r a p h f o r

s i n x l o o k s l i k e ?

b t w : p l e a s e {d o n}^{'} t w r i t e a l l t h i n g s

i n o n l y o n e s i n g l e l i n e! t h a n k s!

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