Does This Limit Chart Indicate a Nonexistent Limit?

[m_s_a]

hi,
find this limit

 

[Dragonfall]

Try: https://www.physicsforums.com/forumdisplay.php?f=156

But to answer your question, $$x\sqrt{A}=\sqrt{x^2A}$$.

 

[HallsofIvy]

Yes, the fact that $x\sqrt{A}= \sqrt{x^2A}$ (and the fact that square root is continuous wherever it is defined) makes the problem simple.

In any case, m s a, this is clearly school work which should have been posted in the "Homework and Coursework" section so I am moving it there.

Also, as you should already know, you MUST show what work you have already done and what thoughts you had about it so we will KNOW what help you need.

 

[CompuChip]

Hi.

I found it.

Can I keep it?

 

[stewartcs]

Try: https://www.physicsforums.com/forumdisplay.php?f=156

But to answer your question, $$x\sqrt{A}=\sqrt{x^2A}$$.

Becareful, this is only true when $$x \geq 0$$.

 

[m_s_a]

Thank you all tried to find the end

but the fee is quite different
These endeavours
Where is the error??

 

[stewartcs]

Thank you all tried to find the end

but the fee is quite different
These endeavours
Where is the error??

I don't understand what you just wrote.

Can you properly state your question please?

CS

 

[m_s_a]

cont.

 

[m_s_a]

chart shows that the end does not exist

 

[m_s_a]

This fee Key
clear that the end does not exist
is not it?

 

[HallsofIvy]

Yes,. the problem is what stewartcs pointed out. If x> 0, then $x\sqrt{4+ 1/x^2}= \sqrt{4x^2+ 1}$ and that has limit 1. But is x< 0, then $x\sqrt{4+ 1/x^2}= -\sqrt{4x^2+ 1}$ and that has limit -1. The "limit from above" is 1 and the "limit from below" is -1. Since those are different, the limit itself does not exist.

 

[m_s_a]

Thank you again
I will return to this subject

 

[stewartcs]

chart shows that the end does not exist

Correct, the limit does not exist.

CS