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 [itex]|\sin x| \leq 1[/itex],
[tex]|x^4 \sin(1/x)| = |x^4| \cdot |\sin(1/x)| \leq |x^4| \cdot 1[/tex]
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: [itex]-1\le \sin(1/x) \le 1[/itex] for all x. Thus [itex]-x^4\le x^4\sin(1/x)≤x^4[/itex]. 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?
The limit of (x^4)(sin1/x) as x approaches 0 is 0. This can be determined by using the Squeeze Theorem and the fact that the sine function is bounded between -1 and 1.
The limit of (x^4)(sin1/x) can be calculated by substituting 0 for x and simplifying the expression. This results in the limit being equal to 0.
The limit of (x^4)(sin1/x) as x approaches 0 exists because the function is continuous and bounded near x=0. This allows us to use the Squeeze Theorem to determine the limit as 0.
No, L'Hopital's rule cannot be used to evaluate the limit of (x^4)(sin1/x) as x approaches 0. This is because L'Hopital's rule can only be used for limits involving indeterminate forms such as 0/0 or ∞/∞, and the given limit does not fall under any of these forms.
Yes, the limit of (x^4)(sin1/x) also exists if the expression is changed to (x^4)(cos1/x). This is because the limit is determined by the behavior of the function near x=0, and both the sine and cosine functions approach 0 as x approaches 0.