# Find this limit

1. Jan 4, 2008

### m_s_a

hi,
find this limit

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2. Jan 4, 2008

3. Jan 4, 2008

### HallsofIvy

Staff Emeritus
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.

4. Jan 4, 2008

### CompuChip

Hi.

I found it.

Can I keep it?

5. Jan 4, 2008

### stewartcs

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

6. Jan 4, 2008

### m_s_a

Thank you all tried to find the end

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

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7. Jan 4, 2008

### stewartcs

I don't understand what you just wrote.

CS

8. Jan 4, 2008

cont.

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9. Jan 4, 2008

### m_s_a

chart shows that the end does not exist

10. Jan 4, 2008

### m_s_a

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

#### Attached Files:

• ###### grafic.zip
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11. Jan 4, 2008

### HallsofIvy

Staff Emeritus
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.

Last edited: Jan 4, 2008
12. Jan 4, 2008

### m_s_a

Thank you again