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I'm trying to understand the true meaning of limits. But a doubt occurred me...
Limits try to tell us the behavior of a function near a particular input, right? Ok...
So:
\lim_{x \to 0^{+}}{1\over x} = \infty
\lim_{x \to \infty}{1\over x} = 0
These two limits are true, right? Ok...
So, by the first limit we can understand that when x tends to 0, 1/x tends to infinity... x will never be zero and 1/x will never be infinity, but for us to get this result in the limit we SUPPOSE that if x would be 0 when 1/x would be infinity. The same thing for the other limit. Am I right? If yes...
So then a limit is a supposition, limits are not "real" (?) because they're a supposition. So what happens to all the things discovered using limits? Are they suppositions too?
Derivatives are defined as a limit when deltaX goes to 0. How can the derivative represent the REAL slope of the line if limits are not real (are only supposition)?
Maybe I was doing a lot of confusion with these concepts, could someone give me a light please?
Thank you,
Rafael Andreatta
Limits try to tell us the behavior of a function near a particular input, right? Ok...
So:
\lim_{x \to 0^{+}}{1\over x} = \infty
\lim_{x \to \infty}{1\over x} = 0
These two limits are true, right? Ok...
So, by the first limit we can understand that when x tends to 0, 1/x tends to infinity... x will never be zero and 1/x will never be infinity, but for us to get this result in the limit we SUPPOSE that if x would be 0 when 1/x would be infinity. The same thing for the other limit. Am I right? If yes...
So then a limit is a supposition, limits are not "real" (?) because they're a supposition. So what happens to all the things discovered using limits? Are they suppositions too?
Derivatives are defined as a limit when deltaX goes to 0. How can the derivative represent the REAL slope of the line if limits are not real (are only supposition)?
Maybe I was doing a lot of confusion with these concepts, could someone give me a light please?
Thank you,
Rafael Andreatta
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