What Are the Limits of Calculus and Why Is b->0- Not Valid?

[Torshi]

Homework Statement

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-

Homework Equations

given above

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?

 

[Dick]

Homework Statement

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-

Homework Equations

given above

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?

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.

 

[Torshi]

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.

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

 

[SammyS]

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

The square root of -1 is i .

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

 

[Dick]

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

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?

 

[Torshi]

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?

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

 

[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-.