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.
I don't understand your question: the xn's and yn's we defined served to show that in any neighborhood of zero f ' has positive and negative values, so it does not keep one single sign in it ==> f isn't increasing or decreasing in no neighborhood of zero. What does this have to do with the actual value of f'(0)?