in Calculus edited by
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Somewhere I found that ( on Google i guess, don't remember the exact site )

$\left | f(x)\right |$  is always continuous .

But $\left | \frac{1}{x-1}\right |$ is discontinuous at       $@$  $x=1$ . 

Why this happens ?

in Calculus edited by
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I think it should be discontinuous at 1 instead of 0 as we are getting the f(0)
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But why property doesn't hold or its not true at all.

look once

@Magma @OneZero 

 

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@HeadShot

WARNING : i am not good at calculus, plz take others opinion also.

So as far as i know mod has no effect on continuity.

Because continuity means their is NO holes in the given function.

now consider a function which has a hole i.e.. its NOT continious, now by making the function positive ( |f(x)| ), the hole still exist, the only difference is that the hole now lies above X-axis instead of below.

so i think the property you read online is not true.       

 

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$\frac{1}{|x-1|}$ graph

we see it from given graph that f(x) is discontinuous from  x = 1 to -1

the graph is continuous at x belongs to (1 to infinite ) U (-infinity to -1) 

therefore it's also discontinuous at x = 0

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