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Old November 1st, 2009, 10:30 PM
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Quote:
Originally Posted by thaopanda View Post
oh wow, I wasn't paying attention. Messed that one up. It's actually:

f(x) :=
x^3, x \leq 0
x^2*sin(\frac{1}{x}), x > 0

sorry about that
To be differentiable, the function needs to be continuous and smooth.


Clearly x^3 is continuous since it is a polynomial.

Also, since x^2 and \sin{\left(\frac{1}{x}\right)} are both continuous for x > 0, so will be their product.

So the only place where this hybrid function might not be continuous is where the function "changes" - so at x = 0. We need to check that the left hand and right hand limits are 0.

Clearly the left hand limit is 0. So we would need to show that \lim_{x \to 0^{+}}x^2\sin{\left(\frac{1}{x}\right)} = 0.
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