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**zhupolongjoe** Let f(x)=x+2x^2*sin(1/x) for x=/=0 and f(0)=0

Prove that f(x) is differentiable everywhere and that f'(0)>0, but there is no neighborhood of 0 on which f is increasing.

I differentiate using the standard rules to get:

f'(x)=-2cos(1/x)+4xsin(1/x)+1 for x=/=0

And when x=0, I look at

lim(x->0) x+2x^2*sin(1/x)/x, which upon factoring, gives that the limit is equal to 1>0, and so f'(0)>0.

The part I can't get is the last part, show there is no neighborhood of 0 on which f is increasing. I know the sin(1/x) term gives wild oscillation, and 1/x is unbounded, but I don't know how to formally show the conclusion that there is no neighborhood of 0 on which f is increasing.

Thanks.