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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 ?

Answer 1

Elias Ackerman

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)#

#color(white)()# 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.

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Answer 2

Lily Anderson

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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Answer from HIX Tutor

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.

Suppose #I# is an interval and function #f:I-&gt;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 ?

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 ?