Final answer:
To verify that lim xcos(1/x) = 0 as x approaches zero, we analyze the behavior of x and cos(1/x). As x becomes very small, 1/x becomes very large, and cos(1/x) oscillates between -1 and 1. Multiplying a bounded oscillating function by a variable approaching zero will result in the limit approaching zero.
Step-by-step explanation:
To verify that lim xcos(1/x) = 0 as x approaches infinity, we need to consider the behavior of the function xcos(1/x). As x grows very large, 1/x approaches 0. Knowing that cos 0 = 1, we consider the entire function. But since x is multiplying the cosine function, as x approaches infinity, cos(1/x) will oscillate between -1 and 1. The product of this bounded oscillation by x, which is approaching zero, will therefore approach 0. This is confirmed by squeezing the cosine function between -1 and 1 and using the fact that x times a number between -1 and 1 as x approaches infinity also approaches 0.
The provided information ranges from various trigonometric concepts to more complex scenarios involving limits and proportions. However, key to our simple limit verification is the understanding that the cosine function yields results between -1 and 1, and as x becomes very large (approaches infinity), the value of 1/x approaches 0, and the cosine of this very small angle is very close to 1. Hence, we see that the function xcos(1/x) is essentially x multiplied by a number that gets infinitely close to 1 as x becomes large. Still, since our focus is on the limit as x approaches zero, the fluctuation in the cosine value becomes irrelevant—the x approaching zero dominates, dragging the value of the entire function to 0.