The discussion revolves around the limits of a function defined as f(b) = 2 - 2√b, specifically examining the limits as b approaches 0 from the positive side and 1 from the negative side, as well as questioning the validity of the limit as b approaches 0 from the negative side.
Participants have provided insights into why the limit as b approaches 0 from the negative side may not be valid, particularly focusing on the undefined nature of the square root of negative numbers in real number contexts. There is ongoing exploration of how to articulate this reasoning clearly.
Participants note that the problem is situated within the realm of real numbers, which influences the discussion about the validity of certain limits and the definitions involved.
Torshi said: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 said: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 said:Thank you! Is it because square root of a negative is "i"
Torshi said:Thank you! Is it because square root of a negative is "i"
Dick said: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?