Thank you for you reply @FactChecker!FactChecker said:
Oh. The reason for not including the point ##\pi/4## is that the rest of the sentence is only about continuity on ##(-\infty, \pi/4)\cup(\pi/4, \infty)##. So there was no need to include ##\pi/4##. It wouldn't have hurt to include it.ChiralSuperfields said:
A piece-wise function is a function that is defined by different expressions or formulas over different intervals of its domain. Each "piece" of the function applies to a specific interval, and the function can have different rules for different parts of its domain.
To determine if a piece-wise function is continuous at a specific point, you need to check three conditions: the function must be defined at that point, the limit of the function as it approaches the point from both the left and the right must exist, and the limit must equal the function's value at that point. Mathematically, this means that the left-hand limit, right-hand limit, and the function's value at the point must all be equal.
Checking the continuity of a piece-wise function at the boundary points is crucial because these are the points where the function switches from one piece to another. Ensuring continuity at these points guarantees that there are no abrupt jumps or breaks in the function, which is essential for many applications in mathematics and applied sciences.
To use limits to show that a piece-wise function is continuous, you need to evaluate the left-hand limit and the right-hand limit at the boundary points where the pieces of the function meet. If these limits exist and are equal to each other and to the function's value at the boundary point, then the function is continuous at that point. This process should be repeated for all relevant boundary points.
Yes, a piece-wise function can be continuous even if it has different formulas on different intervals. The key requirement is that the function must meet the continuity conditions at the boundary points where the formulas change. If the function's value and the limits from both sides of each boundary point are equal, then the function is continuous despite having different expressions on different intervals.