#### Alexander Lorien

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- Thread starter Alexander Lorien
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What is the

Please show us what you have tried and

Please follow the rules of posting in this forum, as enunciated at:

Please share your work/thoughts about this problem.

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Do you know that \(||a|-|b||\le|a-b|~?\)

\(|a|=|a-b+b|\le|a-b|+|b|\) implies \(|a|-|b|\le|a-b|\)

Likewise \(|b|-|a|\le|b-a|=|a-b|\) or \(-|a-b|\le||a|-|b||\le|a-b|\)

So \(\large\left||a|-|b|\right|\le|a-b|\)

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Does this mean that the statement is false??

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Does it mean that the statement is false? Sorry i am confusedDo you know that \(||a|-|b||\le|a-b|~?\)

\(|a|=|a-b+b|\le|a-b|+|b|\) implies \(|a|-|b|\le|a-b|\)

Likewise \(|b|-|a|\le|b-a|=|a-b|\) or \(-|a-b|\le||a|-|b||\le|a-b|\)

So \(\large\left||a|-|b|\right|\le|a-b|\)

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You simply need to sit down with an instructor. You need help.Does it mean that the statement is false? Sorry i am confused

If \(f\) is a contusions at \(x=a\) then if \(c>0\) then \(\exists d>0\) so that \(|x-a|<c\to |f(x)-f(a)|<d\)

BUT \(||f(x)|-|f(a)||\le|f(x)-f(a)| <d\) Does that mean that mean that \(|f|\) is continuous at \(x=a~?\)

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Draw a few functions and see for yourself. You need to see if it is true or not. A few simple drawing will do wonders.Does this mean that the statement is false??