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suju101
 4 years ago
test the differentiability of the function at x=0 F(x)= {x*sin(1/x)if x≠0 and 0 if x=0
suju101
 4 years ago
test the differentiability of the function at x=0 F(x)= {x*sin(1/x)if x≠0 and 0 if x=0

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shankvee\
 4 years ago
Best ResponseYou've already chosen the best response.0It is not differentiable. you can check that the function is continous. Applyinf frist principle \[f \prime(0)=\lim_{h \rightarrow 0}(f(h)f(0))/h=hsin(1/h)/h=\sin1/h\] sin1/h is undefined as h tends to 0 hence it is not differentiable.

suju101
 4 years ago
Best ResponseYou've already chosen the best response.0but can't we say 1/h>infinity

shankvee\
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah since 1/h>infinty the limit sin(1/h) and subsequently f'(0) is undefined.

shankvee\
 4 years ago
Best ResponseYou've already chosen the best response.0If you have x^a sin(1/x) it is differentiable at x=0 only for a>1 for an integer a am not sure for fractional a value.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1The derivative exists at x=0 if the limit of the difference quotient \[ \frac{f(h)  f(0)}{h} = \sin(1/h) \] as h > 0 exists. But it doesn't exist because \[ \lim_{h \rightarrow 0} \sin(1/h) \] doesn't exist. Instead, as h approaches 0, from either above or below, sin(1/h) oscillates faster and faster between 1 and 1. If it had a limit, L say, then we could make sin(1/h) as close as we liked to L by making h sufficiently close to zero. But on the contrary, for any restriction of h close to zero, sin(1/h) assumes infinitely many values between 1 and 1.
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