The discussion revolves around the limits of functions f(x) and g(x) as x approaches 0, particularly focusing on the implications of the inequality |f(x)| ≤ g(x) and the behavior of g(x) at the limit. Participants explore the limit of f(x) when g(x) approaches 0 and when g(x) approaches 3.
The discussion is ongoing, with various interpretations being explored. Some participants have offered guidance on using the squeeze theorem, while others are questioning the assumptions and definitions involved in the problem. There is no explicit consensus on the existence of the limit of f(x) when g(x) approaches 3.
There are mentions of potential typos in the problem statement regarding the limit approaching 0 versus another point, which may affect the analysis. The constraints of the problem, such as the relationship between f(x) and g(x), are also under scrutiny.
lep11 said:Homework Statement
Let |f(x)|≤g(x) for all x∈ Mf∩Mg.
What is limx-->0 f(x) when lim x-->0 g(x)=0?
What is limx-->0 f(x) when lim x-->0 g(x)=3?The Attempt at a Solution
Well, given that|f(x)|≤g(x), lim x-->0 g(x)=0 intuitively implies to me that | limx-->0 f(x) |≤0
AND when lim x-->0 g(x)=3
|limx-->0 f(x)|≤3 ⇔ -3 ≤limx-->0 f(x)≤3
Is my reasoning correct?
How?pasmith said:Can you turn your intuition into a rigorous proof that \lim_{x \to 0} f(x) = 0 in this case?
No. So in that case we cannot say anythin about \lim_{x \to 0} f(x) since it won't necessarily even exist?pasmith said:Must \lim_{x \to 0} f(x) necessarily exist when \lim_{x \to 0} g(x) > 0? Consider the case <br /> f(x) = \begin{cases} -1 & x < 0, \\ 0 & x = 0, \\ 1 & x > 0, \end{cases}\qquad g(x) = 3.
lep11 said:ahh there is a small typo in the problem statement.. it should be x approaches a, not 0
Isn't it the limit as x approaches a?So for the second part, if ##|f(x)| \le g(x)## and ##\lim_{x \to a} g(x) = 3##, is it necessarily true that ##\lim_{x \to a} f(x) = 0##?lep11 said:I used the squeeze theorem to prove that limx→0f(x)=0
Yes it is, I'm sorryMark44 said:Isn't it the limit as x approaches a?
No, because limx→a f(x) won't necessarily exist when lim x-->a g(x)=3.Mark44 said:So for the second part, if ##|f(x)| \le g(x)## and ##\lim_{x \to a} g(x) = 3##, is it necessarily true that ##\lim_{x \to a} f(x) = 0##?
lep11 said:I used the squeeze theorem to prove that limx→0f(x)=0
Mark44 said:So for the second part, if ##|f(x)| \le g(x)## and ##\lim_{x \to a} g(x) = 3##, is it necessarily true that ##\lim_{x \to a} f(x) = 0##?
OK, I wasn't sure whether you were using the squeeze theorem for the second part.lep11 said:No, because limx→a f(x) won't necessarily exist when lim x-->a g(x)=3.