- #1
kbrono
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Let (a, b) be an open interval in R, and p a point of (a, b). Let f be a real-valued function defined on all of (a, b) except possibly at p. We then say that the limit of f as x approaches p is L if and only if, for every real ε > 0 there exists a real δ > 0 such that 0 < | x − p | < δ and x ∈ (a, b) implies | f(x) − L | < εI have absolutely no idea how to go about this proof.
Here's my first part attempt
ep = epsilon
d= delta
Suppose f(x) is defined near p and for every ep>0 there exists a d>0 such that for every x in R with 0<|x-p|<d, |f(x)-L|<ep. Because f(x) is near p that means that there is some d>0 such that 0<|x-p|<d and x is to be near p that means there is some d>0 such that 0<|x-p|<d and x is in the domain of f(x). Since we know that x in the domain of f(x) and 0<|x-p|<d for some d>0 then we have |f(x)-L|<ep. Which is the definition that the lim f(x) as x->p = L.
Here's my first part attempt
ep = epsilon
d= delta
Suppose f(x) is defined near p and for every ep>0 there exists a d>0 such that for every x in R with 0<|x-p|<d, |f(x)-L|<ep. Because f(x) is near p that means that there is some d>0 such that 0<|x-p|<d and x is to be near p that means there is some d>0 such that 0<|x-p|<d and x is in the domain of f(x). Since we know that x in the domain of f(x) and 0<|x-p|<d for some d>0 then we have |f(x)-L|<ep. Which is the definition that the lim f(x) as x->p = L.
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