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I have the following problem : Let $f:(0,\infty)\rightarrow \mathbb R$ be an arbitrary function satisfying the hypothesis, $\lim_{x \rightarrow 0} x(f(x)-1)=0$. Show that $\lim_{x \rightarrow 0 } f(x)$ exists.

The problem has also other parts and 2 other hypothses on $f$. But the solution starts like this: by hypothesis one it follows $\lim_{x \rightarrow 0} xf(x)=1$... So it doesn't use the other two hypothesis ( why would he explicitly state that the result follows from hyp. one if the others are also used?).

But than $ \ \sin(\frac{1}{x}):=f(x) \ $ would be a conter example.

Can somone say if my counterexample is right?I can't see the error. And I'm pretty sure he doesn't use the other hypotheses since he doesn't state them in the problem.

here are the other two conditions on $f$ (wich i didn't use in the counterexample by the reasons explained above)

two : $f(1)=2$

three: $(x+2)f(x+2)-2(x+1)f(x+1)+xf(x)=0$

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your counter example is right – Un Chien Andalou Nov 5 '13 at 15:08
Your counterexample is correct. (Beaten!) – Ahaan S. Rungta Nov 5 '13 at 15:09
thanks for the confirmations! – sigmatau Nov 5 '13 at 15:27
You're welcome! Thanks for the accept. :) – Ahaan S. Rungta Nov 5 '13 at 15:34
up vote 1 down vote accepted

Your counterexample is correct. (Converting to an answer.)

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