# Computing these limits?

1. Jan 28, 2013

### Torshi

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

Just checking if these are right?

f(b) = 2-2√b
compute limit in are as:
b-> 0+
b->1-
Explain in a brief sentence why it does not make sense to compute a limit as b->0-

2. Relevant equations
given above

3. The attempt at a solution
lim b->0+ (2-2√b) = 2
lim b->1- (2-2√b) = 0
Lim b->0- (2-2√b) = ? The question says it doesn't make sense to do so. Is it not 2?

2. Jan 28, 2013

### Dick

Fine for the first two. For the last if b->0- then b is negative. If you are working in the real numbers then taking the square root of a negative number should make you feel odd at least. Why? This is another example of where blind plugging without thinking is not the best idea.

3. Jan 28, 2013

### Torshi

Thank you! Is it because square root of a negative is "i"

4. Jan 28, 2013

### SammyS

Staff Emeritus
The square root of -1 is i .

$\displaystyle \sqrt{-|a|}=i\cdot\sqrt{|a|}$

5. Jan 29, 2013

### Dick

The square root of -1 is also -i, but main point is that if you are working with real numbers and limits the square root of a negative number is simply undefined. It's not real. I think that's what the are expecting you to say. Can you tell me why there is no real number whose square is negative?

6. Jan 29, 2013

### Torshi

I should of been more clear in terms of stating it being undefined. Sorry

7. Jan 29, 2013

### Dick

I wasn't looking for an apology and there is none needed. I was just looking you to say how you would explain that there was no limit b->0-.