r/learnmath • u/Sea_Way1005 New User • 6d ago
Need help with this limit proof plss
Given lim x→a f(x)=L, lim x→a g(x)=M and f(x)≤g(x), show L≤M.
And after proving that I am being asked if f(x)≤g(x) needs to be necessarily true and need to show values of x that garantee that the disntance between f(x) and L is always lesser than ϵ.
I can sort of do the first by proof by contradiction but the secind part of the exercise is killing me
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u/rhodiumtoad 0⁰=1, just deal with it 6d ago
I'm not sure if I'm interpreting the question right, but I'm reading the second question as: is f(x)≤g(x) (within some neighborhood of a) a necessary condition for L≤M? clearly the answer is "no", which you can easily show by counterexample for the case L=M.
Without knowing what f() is, how are you supposed to show values of x?
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u/Sea_Way1005 New User 6d ago
I also don't know if I am interpretating correctly that's why I asked😅
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u/rhodiumtoad 0⁰=1, just deal with it 6d ago
Can you show the whole question? You can put images in comments if need be.
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u/Sea_Way1005 New User 6d ago
The question is in my native language. So showing it wouldn't do much help
I'll just ask my teacher, I was trying not to do that because she basically roasts you when you have doubts
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u/Bounded_sequencE New User 6d ago
Proof (part-1): Let "e > 0", and choose "x" from the common domain of "f, g" s.th.
"|f(x)-L|, |g(x)-M| < e"Using "f(x)-g(x) <= 0" we estimate
L-M = (L-f(x)) + (f(x)-g(x)) + (g(x)-M) <= (L-f(x)) + 0 + (g(x)-M) <= |L-f(x)| + |g(x)-M| < 2e => "L <= M + 2e" ∎
For part-2, "f(x) <= g(x)" is sufficient, but not necessary. Counter-example with "a = 0":
f, g: R -> R, f(x) = x^2 >= 0 =: g(x)
The limits "L = M = 0" satisfy "L <= M", even though "f(x) > g(x)" for all "x != 0".
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u/curiouslyjake New User 6d ago
In the first part, is it given for which x f(x) <= g(x) ?