The limit of the function (x^4)(sin(1/x)) as x approaches 0 exists and is equal to 0. This conclusion is reached by applying the Squeeze Theorem, which utilizes the bounded nature of the sine function, specifically that -1 ≤ sin(1/x) ≤ 1. Despite the individual limits of sin(1/x) being undefined as x approaches 0, the product with x^4, which approaches 0, results in the overall limit being 0.
PREREQUISITESStudents studying calculus, particularly those focusing on limits and trigonometric functions, as well as educators looking for examples of applying the Squeeze Theorem in limit problems.
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?