why does sin(1/x) diverge, while sin(1/x^2) converge?
To understand why the function sin(1/x) diverges while sin(1/x^2) converges, we need to analyze the behavior of these functions as x approaches zero.
Let's start with the function sin(1/x). As x approaches zero, the argument of the sine function becomes increasingly large and oscillates more frequently. This means that as x gets closer to zero, the sine function will oscillate rapidly between -1 and 1. The amplitude of these oscillations does not diminish, leading to a divergence.
On the other hand, let's consider the function sin(1/x^2). As x approaches zero, the argument of the sine function becomes increasingly large, but since it is squared, the values are crowding towards infinity much faster. As a result, these oscillations become more closely spaced, effectively "squeezing" the function between two horizontal lines (the values of -1 and 1). This behavior limits the oscillations and prevents them from diverging, resulting in convergence.
To further confirm this, we can use mathematical analysis techniques like limits:
1. For sin(1/x): The limit as x approaches zero of sin(1/x) does not exist. This divergence can be proved by considering different sequences that approach zero, and showing that the function does not have a unique limit.
2. For sin(1/x^2): The limit as x approaches zero of sin(1/x^2) is zero. By using the squeeze theorem, we can show that -1 ≤ sin(1/x^2) ≤ 1 for all x approaching zero, implying that the limit exists and equals zero.
In summary, the key difference between sin(1/x) and sin(1/x^2) lies in the behavior of their arguments as x approaches zero. The inverse relationship of the argument in the latter function results in convergence, while the direct relationship in the former leads to divergence.