What is the limit of a function under a square root?

[houssamxd]

Homework Statement

lim x-> 2+ f(x)=sqrt(4-x^2)whats the value of the following function??

Homework Equations

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

 

[fzero]

Can you explain the steps that lead to your result?

 

[Mark44]

Homework Statement

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


whats the value of the following function??

Homework Equations

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

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.
$$\lim{x \to 2^-}\sqrt{4 - x^2}$$

 

[houssamxd]

Can you explain the steps that lead to your result?

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

 

[houssamxd]

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.
$$\lim{x \to 2^-}\sqrt{4 - x^2}$$

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

 

[Mark44]

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

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

 

[houssamxd]

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

so I am right
it doesn't exist

 

[Mark44]

Yes. The limit doesn't exist.

Your first post in this thread was confusing to me.

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

which is the correct answer

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.

 

[houssamxd]

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.

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

 

[Mark44]

sorry about that
but anyway thanks for your ttime and help

You're welcome!

i owe you one

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.

 

[Ray Vickson]

so I am right
it doesn't exist

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
$$4 - x^2 = -(x^2 - 4) = (x^2 - 4) e^{\pm i \pi} \, \Longrightarrow<br /> \sqrt{4 - x^2} = \sqrt{x^2 - 4}\: e^{\pm i \pi/2} = \pm\, i \sqrt{x^2 - 4}.$$
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.