Could anyone take a look at question 5 from PART-II of DSE 2009?
5. Consider the function
f(x)= x^2. sin(1/x) if x=/ 0
=0 if x=0
Then the following is true about the derivative of f :
a) 1 f )0(' = − and ) f (' x is continuous at x = 0.
b) f )0(' = −1 and ) f (' x is discontinuous at x = 0.
c) f )0(' = 0 and ) f (' x is discontinuous at x = 0.
d) f (' x)is not defined at x = 0.
Basically, I don't know how to solve the LHL and RHL. Could anyone show the steps for it?
f(x) is continuous at x=0. Check with LHL=RHL = value of the function.
f(x) is clearly differentiable. Check with left hand differentiability and right hand differentiability at x=0 to see that f(x) doesn't have a "kink" at x=0.
f'(x)= 2xsin(1/x)-cos(1/x) when x ≠ 0
= 0 when x=0
now we need to check the continuity of g(x)=f'(x)
the limit of g(x) is definitely ≠ 0 (infact it doesn't even exist) when x tends to zero. hence g(x) which is nothing but f'(x) is discontinuous at x=0.
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