Suppose $I$ is an interval and function $f:I->R$ and $x in I$ . Is it true that $f(x)=1/x$ is not bounded function for $I=(0,1)$ ?. How do we prove that ?

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Answer 1 · 2017-04-11T06:22:28

See explanation...

Given:

$f(x) = 1/x$

$I = (0, 1)$

Then for any $N > 0$ we have:

$1/(N+1) in (0, 1)$

$f(1/(N+1)) = N+1 > N$

So $f(x)$ is unbounded on $(0, 1)$

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Comments

This example shows the necessity of the condition that the interval be closed in the boundedness theorem:

A continuous function defined on a closed interval is bounded in that interval.

Suppose II is an interval and function f:I->Rf:I->R and x in Ix in I . Is it true that f(x)=1/xf(x)=1/x is not bounded function for I=(0,1)I=(0,1) ?. How do we prove that ?