How to show that f is not continuous at 0?
That means to show that f is not continuous at 0, it is sufficient to exhibit one sequence (x n) which converges to 0 for which the sequence (f (x n) does not converge to f (0)=1. So the suggestion was take x n =1/n.
Is the function f (x) continuous everywhere?
Consider f:[?1,3]?R, f(x)= ! 2x, ?1? x ?1 3?x,1
Is f uniformly continuous?
If f (x)=x and the domain is R, then f is uniformly continuous. (We can take delta=epsilon.) The range of this f is all of R, an unbounded set. If f (x)=sin (1/x) and the domain in (0,1), then f is not uniformly continuous.
Is F a continuous curve?
If f is continuous on [ 1, 2] (i.e., its graph can be sketched as a continuous curve from ( 1, ? 10) to ( 2, 5)) then we know intuitively that somewhere on [ 1, 2] f must be equal to ? 9, and ? 8, and ? 7, ? 6, …, 0, 1 / 2, etc. In short, f takes on all intermediate values between ? 10 and 5.