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Suppose $f(x)$ is defined and $\mathbf{continuous}$ on $(-1, 1)$. Given 3 statements:

(i) If the limit $\lim_{x \to 1-0} f(x)$ exists and is finite then the function achieves either the largest or the least value on $(-1, 1)$

(ii) If $f$ achieves the largest and the least its values on $(-1, 1)$ then the limit $\lim_{x \to 1-0}f(x)$ exists and is finite

(iii) If limits $\lim_{x \to 1-0}f(x)$ and $\lim_{x \to -1+0}f(x)$ exist and are finite then the function achieves the largest and the least its values on $(-1, 1)$

And it's argued that all statements are $\mathbf{false}$. I've failed to think up counter examples and also I don't understand why these statements are false, for example as for (ii) we know that the function achieves its the largest and the least values and is continuous hence it's bounded (am I right?) therefore such limit from (ii) must exist (because if it's equal to $+\infty$ or $-\infty$ then $f(x)$ isn't bounded at the left-side neighbourhood of the point $x=1$ ). Also I am able to give the similar wrong "proofs" for (i) and (iii) but I think it will be redundant.

Could you please help me to think up counter examples or indicate the mistakes in my considerations? I would be grateful for any hints. Thanks in advance.

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  • $\begingroup$ I would gladly give you some counterexamples, but I don't understand your notation "$\lim_{x \to 1-0}$", can you explain it to me? $\endgroup$ – Netchaiev Feb 11 '18 at 10:33
  • $\begingroup$ @Netchaiev it means one-sided limit $\endgroup$ – D F Feb 11 '18 at 10:38
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For (i) and for (iii), just take $f(x)=x$.

For (ii), take $f(x)=\sin\left(\frac1{x-1}\right)$. Both the largest and the least values are attained in $(-1,1)$, but $\lim_{x\to1^-}f(x)$ does not exist.

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