Exploring the Limit of sin(√x1) - sin(√x) as x Approaches Infinity

In this article, we delve into the intricacies of determining the limit of the expression sin(sqrt{x_1}) - sin(sqrt{x}) as x approaches infinity. This involves a meticulous analysis of the behavior of the sine function and its application of fundamental theorems from calculus.

The Fundamental Theorem and Squeeze Theorem

The behavior of the given expression can be understood through the lens of the Mean Value Theorem (MVT) in calculus. The MVT states that if a function f is continuous on [a, b] and differentiable on (a, b), then there exists a point c in (a, b) such that the derivative of f at c is equal to the average rate of change over the interval. In our case, we apply this theorem to the function f(t) sin(sqrt{t}) over the interval [x, x_1].

Mean Value Theorem Application

Using the MVT, we can write:

Let (varphi(x) sin(sqrt{x_1}) - sin(sqrt{x})) According to MVT, there exists a (theta) in the interval [x, x_1] such that:

(varphi(x) frac{cos(theta)}{2sqrt{theta}})

Now, knowing that (costheta leq 1) and (frac{1}{2sqrt{theta}} leq frac{1}{2sqrt{x}}), we can derive the inequality:

(frac{1}{2sqrt{x}} geq varphi(x) geq -frac{1}{2sqrt{x}})

As (x rightarrow infty), both (frac{1}{2sqrt{x}} rightarrow 0) and (-frac{1}{2sqrt{x}} rightarrow 0). Thus, by the Squeeze Theorem, we have:

(lim_{x rightarrow infty} varphi(x) 0)

Alternative Simplification Using Taylor Series

Another approach is to use Taylor series expansion. When xrightarrowinfty, we observe that (sqrt{x_1}) is very close to (sqrt{x}). This implies that the difference (sqrt{x_1} - sqrt{x}) becomes very small, which makes the entire function approach zero. Formally, as (xrightarrowinfty), we can approximate the numerator as 1 and the denominator as an infinitely large value, leading to an expression approaching zero.

Taylor Series Expansion

Let's apply the Taylor series approximation:

Using the identity sin(C) - sin(D) 2cos(frac{C D}{2})sin(frac{C-D}{2}), let (C sqrt{x_1}) and (D sqrt{x}):

( sin(sqrt{x_1}) - sin(sqrt{x}) 2cos(frac{sqrt{x_1} sqrt{x}}{2})sin(frac{sqrt{x_1} - sqrt{x}}{2}) )

Since the denominator (cos(frac{sqrt{x_1} sqrt{x}}{2})) is bounded (less than or equal to 1), and the numerator (sin(frac{sqrt{x_1} - sqrt{x}}{2})) converges to 0 as (x rightarrow infty) (due to the oscillatory nature of the sine function with a small argument), we conclude that:

( (sin(sqrt{x_1}) - sin(sqrt{x})) 2cos(frac{sqrt{x_1} sqrt{x}}{2})sin(frac{sqrt{x_1} - sqrt{x}}{2}) rightarrow 0)

Conclusion

The limit of sin(sqrt{x_1}) - sin(sqrt{x}) as x approaches infinity is 0. This conclusion is derived through both the application of the Mean Value Theorem and Taylor series approximation, both of which lead to the same result. The essence of this exploration lies in understanding the behavior of the sine function and employing theoretical tools to analyze its limit.