# Yes or no, Can you square root a zero

1. Mar 5, 2007

### venger

Yes or no, Can you square root a zero. This is for a limits question.

2. Mar 5, 2007

### Doodan

yes and the answer you get is 0 and u dont have to worry about +/- obviously

3. Mar 5, 2007

### venger

Does anyone have any good sites that can help me with limits

4. Mar 5, 2007

### turdferguson

If youre asking about the limit of $$\sqrt{x}$$, you can only approach it from the right. Because the limit point is unapproachable from the left, the limit does not exist. Remember, its not a polynomial so you cant just plug it in

5. Mar 6, 2007

### Gib Z

Ahh it is approachable from the left, but is imaginary. So the limit from the left exists, but isnt the same as the limit from the right, so the general limit does not exist.

6. Mar 6, 2007

### Dick

$$\sqrt{x}$$ is DEFINED to be the POSITIVE square root. In this sense it is undefined for x<0 (neither square root is 'positive').To consider the limit from the left you would have to make a choice of definition of square root that would select one of the square roots for x<0 as well. No matter which way you do this you would still get 0 as the left hand limit.

7. Mar 6, 2007

### turdferguson

If you make the rather basic assumption that the question was posed over reals in an R^2 Cartesian plane, than imaginary numbers are not up for discussion. Furthermore, the concept of order (less than, greater than) dissolves in the complex number system, which makes delta-epsilon pretty hard to do

8. Mar 7, 2007

### Dick

Order may go, but you still have the notion of |x|<epsilon. You can still discuss limits and continuity perfectly well.

9. Apr 26, 2009

### k-hursh

I am also eager for an answer to this question though it is for different reasons. Due to a poor choice in essay questions :) , I am trying to calculate the Doppler Shift in light if one were to be traveling in a car at the speed of light and turn on their headlights. This, I think, should be calculated with the Relativistic Doppler Shift Formula. In the end this leaves me with the square root of a quotient, of which, the denominator is velocity over c plus negative one, or zero. This is not the final quantity of the solution but rather one quantity involved. Does this leave me with an answer that is undefined( making it not possible to have an answer?) or do I have the square root of zero equaling zero therefore simply plugging zero into the equation? PLEASE HELP :) thanks

10. Apr 26, 2009

### Dick

The square root of zero is zero. No doubt about that. If you get a formula with zero in the denominator you have a problem. That's not defined. The v is the doppler shift formula is a relative v between a source and an observer. Who's the observer? In a physical situation it's not possible for v to reach c. You can only ask what happens as v approaches c.

11. Apr 26, 2009

### k-hursh

So I do not hijack this thread I think I should post my full inquiry on a fresh thread.

12. Apr 26, 2009

### Dick

Good idea. You'll get more notice that way as well.