Is the function F(x)= the square root of 1-x^2 continuous at x = 1? Give the reason for your answer.
what is f(x) at x=.99 and at 1.01 ? Are they approaching the same limit?
Yeah?
If they are approaching the same limit and there is a value at f(1) then f(x) is continuous at that point.
To determine if the function f(x) = √(1-x^2) is continuous at x = 1, we can follow a step-by-step approach.
1. First, let's check if the function is defined at x = 1. Since the function involves taking the square root of a term containing 1 - x^2, we need to ensure that the term 1 - x^2 is non-negative at x = 1.
Plug x = 1 into the expression: f(1) = √(1 - 1^2) = √0 = 0.
Since 1 - 1^2 = 1 - 1 = 0, the expression becomes √0, which evaluates to 0. Therefore, the function is defined at x = 1.
2. Next, we need to determine if the function has a limit as x approaches 1. This involves examining the behavior of the function as x gets arbitrarily close to 1, from both the left and right sides.
Let's consider the left-hand limit as x approaches 1: lim(x→1-) √(1 - x^2). This means we evaluate the function as x approaches 1 from values that are slightly less (to the left of 1).
Substituting a value slightly less than 1, such as x = 0.9, into the function: f(0.9) = √(1 - 0.9^2) = √(1 - 0.81) = √0.19 ≈ 0.4359.
Now, let's consider the right-hand limit as x approaches 1: lim(x→1+) √(1 - x^2). This means we evaluate the function as x approaches 1 from values that are slightly greater (to the right of 1).
Substituting a value slightly greater than 1, such as x = 1.1, into the function: f(1.1) = √(1 - 1.1^2) = √(1 - 1.21) = √(-0.21).
At this point, we encounter an issue. The expression inside the square root becomes negative when evaluating the right-hand limit. Since the square root function only provides real values for non-negative arguments, we cannot evaluate √(-0.21), which means the right-hand limit does not exist.
3. Since the left-hand limit exists and is finite (equal to approximately 0.4359), while the right-hand limit does not exist, the overall limit as x approaches 1 does not exist. Therefore, the function is not continuous at x = 1.
In conclusion, the function f(x) = √(1 - x^2) is not continuous at x = 1 due to the non-existence of the limit as x approaches 1.