# 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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Given:

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.

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Yes, it is true that the function ( f(x) = \frac{1}{x} ) is not bounded on the interval ( I = (0,1) ). To prove this, we can show that as ( x ) approaches 0 or 1, ( f(x) ) becomes arbitrarily large.

As ( x ) approaches 0 from the right side, ( f(x) ) approaches positive infinity. Similarly, as ( x ) approaches 1 from the left side, ( f(x) ) approaches positive infinity. Therefore, ( f(x) ) is unbounded on the interval ( I = (0,1) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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