The discussion revolves around evaluating the limit of the expression (x^4)(sin(1/x)) as x approaches 0. Participants are exploring the behavior of this limit, particularly focusing on the oscillatory nature of sin(1/x) and its implications for the limit's existence.
Participants are actively engaging with different interpretations of the limit. Some have suggested that the limit exists and is zero based on bounding arguments, while others are still questioning the assumptions and reasoning behind the limit's evaluation.
There is a discussion about the bounded nature of sin(1/x) and its implications for the product with x^4. The conversation also touches on the use of the squeeze theorem as a potential approach to understanding the limit.
awkward said:The limit as x approaches 0 of sin(1/x) does not exist, so you can't use the usual theorems about products (or quotients) of limits.
But since |\sin x| \leq 1,
|x^4 \sin(1/x)| = |x^4| \cdot |\sin(1/x)| \leq |x^4| \cdot 1
so the limit does exist and is zero.
The problem isn't asking about 1/x. It's asking about sin(1/x), and in particular, the it is asking about the product of x4 with the sine of 1/x. You are correct in the sense that sin(1/x) is undefined as x→0. It is however bounded: -1\le \sin(1/x) \le 1 for all x. Thus -x^4\le x^4\sin(1/x)≤x^4. What does the squeeze theorem say about this as x→0?smeiste said:Since as 1/x → 0+, the limit is +∞ but when 1/x → 0-, the limit is -∞. Since these two-sided limits don't match I thought the limit could not be calculated. apparently this is incorrect. any thoughts?