The definition of continuity in \(\displaystyle \mathbb{R}\) is given in Stephen Abbott's book: Understanding Analysis, as follows:Cbarker1 said:
Thanks for the help, HallsofIvy ...HallsofIvy said:
The definition of continuity for a function is that the function's output values change smoothly and gradually as the input values change. In other words, there are no sudden jumps or breaks in the graph of the function.
To prove that Arcsin x is continuous, we need to show that it satisfies the three conditions of continuity: 1) the function is defined at the point, 2) the limit of the function at that point exists, and 3) the limit is equal to the value of the function at that point. Using the definition of Arcsin x and the properties of limits, we can show that all three conditions are met, thus proving that Arcsin x is continuous.
The domain of Arcsin x is [-1, 1], which means that the input values (x) must be between -1 and 1. The range of Arcsin x is [-π/2, π/2], which means that the output values (y) must be between -π/2 and π/2.
Yes, at x = 1, Arcsin x is not continuous because the limit of the function at x = 1 does not exist. The left-hand limit is π/2 and the right-hand limit is -π/2, which are not equal. This means that there is a sudden jump or break in the graph of the function at x = 1.
The continuity of Arcsin x is directly related to the continuity of its inverse function, Sin x. Since Arcsin x is continuous, it means that Sin x is also continuous, which allows us to use the properties of continuity to solve equations involving Sin x. Additionally, the continuity of Arcsin x ensures that the inverse function, Sin x, exists and is also continuous.
