| T O P I C R E V I E W |
| Manu |
Posted - 04/13/2012 : 21:35:53
f(0) = 0
f(0+) = f(0-) = 0 as f(x) = x2 sin(1/x)
So, f(x) is continous at x=0.
Now, as it is continous, for its differentiability we cant differentiate f(x) and then check for its continuity.
So, we have,
d/dx (f(x)) for x not equal to 0 as :
d/dx (f(x)) = 2x sin(1/x) + x2 cos (1/x) (-1/x2) = 2x sin(1/x) - cos(1/x) , for all x not equal to 0.
But,
for f'(0) it doest not exist, as cos(1/x) does not exist for x=0.
Am I correct. Any help would be good. |
| 1 L A T E S T R E P L I E S (Newest First) |
| someguy |
Posted - 04/13/2012 : 21:46:02
Hi Manu, to find f'(0), use the definition of a derivative.
f'(0) = limit[x -> 0] (f(x) - f(0))/(x - 0)
It may help to recall that sin(1/x) is bounded between -1 and 1 for all x not equal to 0. |
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