# Function under square root

1. Nov 4, 2013

### houssamxd

1. The problem statement, all variables and given/known data

lim x-> 2+ f(x)=sqrt(4-x^2)

whats the value of the following function??

2. Relevant equations

3. The attempt at a solution

i tried and got the answer as does not exist
but some people got it as 0

which is the correct answer

2. Nov 4, 2013

### fzero

Can you explain the steps that lead to your result?

3. Nov 4, 2013

### Staff: Mentor

f(2) = 0, but if x > 2, then the function is undefined. This means that the limit you showed does not exist.

This limit does exist, however, and is equal to 0.
$$\lim{x \to 2^-}\sqrt{4 - x^2}$$

4. Nov 4, 2013

### houssamxd

i assumed that x=2.1
as we approach 2 from the right side

the when i put it under the when we put it under the sqaure and subtract we get
sprt(-0.1)

which is undefined

5. Nov 4, 2013

### houssamxd

but which is correct
is my answer when x->2+ correct

6. Nov 4, 2013

### Staff: Mentor

No. The limit as you wrote it doesn't exist.

7. Nov 4, 2013

### houssamxd

so im right
it doesnt exist

8. Nov 4, 2013

### Staff: Mentor

Yes. The limit doesn't exist.

Your first post in this thread was confusing to me.
I didn't understand that you were asking a question since you didn't end it with a question mark (?). I interpreted what you wrote as saying that 0 was the correct answer.

9. Nov 4, 2013

### houssamxd

sorry about that
but anyway thanks for your ttime and help
i owe you one

10. Nov 4, 2013

### Staff: Mentor

You're welcome!
That's OK. It's what we do here, and we enjoy doing it, as long as you make a reasonable effort when you post a question.

11. Nov 4, 2013

### Ray Vickson

In the real plane the limit does not exist because the function itself does not exist when x > 2. However, in the complex plane the function does exist, and the limit is 0. This is because for x > 2 we have
$$4 - x^2 = -(x^2 - 4) = (x^2 - 4) e^{\pm i \pi} \, \Longrightarrow \sqrt{4 - x^2} = \sqrt{x^2 - 4}\: e^{\pm i \pi/2} = \pm\, i \sqrt{x^2 - 4}.$$
The principal square root uses "+i", but the other is also a solution of z^2 = 4 - x^2. Anyway, no matter which square root you choose, it goes to zero along the imaginary axis in the complex plane, so the limit is zero.

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