Prove that the functions f(x)=x+2, if x=<1 and f(x)=x^2. if x>1 are discontinuous.
The function f(x) is defined such that f(x) = x + 2 , if x <= 1 and f(x) = x^2, if x > 1.
At the point x = 1, if we approach from the left
lim x--> 1- [ f(x)] = 1 + 2 = 3.
If we approach from the right,
lim x--> 1+ [f(x)] = 1^2 = 1
The value of lim x--> 1- [ f(x)] is not equal to lim x--> 1+ [f(x)].
Therefore the function is discontinuous.
We notice that the point where we have to verify the continuity of the function is x = 1.
The function is continuous over the ranges (-infinite ; 1) and (1 ; +infinite).
Now, we'll verify the continuity of the function by evaluating the lateral limits of the function.
For x<1, we'll calculate the limit of the function f(x) = x+2.
lim f(x) = lim (x+2) = 1 + 2 = 3
For x>1, we'll calculate the limit of the function f(x) = x^2.
lim f(x) = lim x^2 = 1^2 = 1
The values of the lateral limits are finite but they are different, so the function is discontinuous in the points x = 1.