Proving Sin(x²) Continuity, Boundedness, And Non-Uniform Continuity

Hey guys! Let's dive into a classic calculus problem that explores the properties of the function f(x) = sin(x²). We're going to demonstrate that this function is continuous and bounded across the entire real number line (from negative infinity to positive infinity), but it's not uniformly continuous. This is a great example to understand the subtle differences between continuity and uniform continuity, and how boundedness plays a role. Buckle up, because we're about to get mathematical!

Showing f(x) = sin(x²) is Continuous

Alright, first things first, let's tackle continuity. A function is continuous at a point if, as you approach that point, the function's value gets closer and closer to the actual value at that point. For f(x) = sin(x²), we can break this down into a composition of two functions: the squaring function g(x) = x² and the sine function h(x) = sin(x). We know a few key things about these component functions, which is super useful.

The function g(x) = x² is a polynomial, and polynomials are continuous everywhere. This means that for any real number x, you can find the value of x² without any jumps or breaks. It’s a smooth curve, always moving in a predictable way. Then we have h(x) = sin(x). The sine function is also continuous everywhere. The graph of sin(x) wiggles smoothly up and down, and at any given point you can zoom in and find the value of the function without any interruptions.

Now, a fundamental theorem in calculus states that the composition of continuous functions is also continuous. In other words, if you have a continuous function inside another continuous function, the result is continuous as well. Since g(x) = x² is continuous and h(x) = sin(x) is continuous, their composition, f(x) = h(g(x)) = sin(x²), must also be continuous. This is a really important tool to understand. To clarify, since both x² and sin(x) are continuous, then sin(x²) is continuous. You can literally think about it as, no matter where you are on the number line, you can plug in any x value into sin(x²), and the function will produce a defined output, without any sudden jumps or holes in the graph. Therefore, f(x) = sin(x²) is continuous everywhere on the interval (-∞, +∞). Think of it like this: the squaring of x doesn't introduce any discontinuities, and the sine function smoothly takes the result of the squaring and maps it to a value between -1 and 1. Pretty neat, huh?

Because the function is continuous, this means that for any value on the real line, we can approach the function's value by taking values very close to the x-value. Because there are no breaks in the curve, we can always find values that fit this criteria, and thus the function is continuous.

Demonstrating f(x) = sin(x²) is Bounded

Next up, let's prove that f(x) = sin(x²) is bounded. A function is bounded if its values stay within a certain range. Think of it as the function's graph being confined between two horizontal lines. In simpler terms, there's an upper limit and a lower limit that the function's output can never exceed. Now, we're dealing with the sine function, and the sine function is known for its oscillating behavior, which is awesome for being bounded.

We know that the sine function, sin(u), always outputs values between -1 and 1, inclusive. Mathematically, this is expressed as: -1 ≤ sin(u) ≤ 1. Here, u can be any real number. In our case, u = x². So, no matter what value x² takes (and it can take any non-negative value), the sine function will always output a value between -1 and 1. Since x² can be any value, sin(x²) will only ever produce values from -1 to 1. This is the definition of boundedness.

Since f(x) = sin(x²) is the composition of x² and the sine function, the output of x² is fed into the sine function. However, the sine function will always be bounded, even when the input is a large number. Consequently, the output will never exceed these boundaries. Consequently, since the sine function's range is limited to [-1, 1], f(x) = sin(x²) is also bounded. Therefore, no matter how big x gets, the function's value remains within this range, proving that f(x) is bounded. This is in contrast to a function like x², which grows unboundedly as x increases. So, f(x) = sin(x²) is bounded on the interval (-∞, +∞).

Proving f(x) = sin(x²) is Not Uniformly Continuous

Now, this is where things get a bit more interesting! We know that continuity, as we discussed previously, implies that the function is continuous at every individual point. However, uniform continuity is a stronger condition. A function is uniformly continuous on an interval if, for any given ε > 0, there exists a δ > 0 such that for any two points x and y in the interval, if the distance between x and y is less than δ, then the distance between f(x) and f(y) is less than ε. Sounds like a mouthful, right? Let's break it down.

In essence, uniform continuity means that you can make the function's values arbitrarily close to each other (ε) by ensuring that the inputs are also sufficiently close to each other (δ), and this works everywhere in the interval. The key thing here is that δ must depend only on ε and not on the specific location within the interval. For a function that's not uniformly continuous, the closeness of the function's values doesn't depend only on the input difference, but also on where it is on the graph.

Let's get into the specifics of f(x) = sin(x²) not being uniformly continuous. We will approach this using a proof by contradiction. The strategy is to find two sequences of points, xₙ and yₙ, where the distance between xₙ and yₙ approaches zero, but the distance between f(xₙ) and f(yₙ) does not approach zero. That is, the values of the function do not get arbitrarily close to each other as the inputs get close. The function will