In the square root function f is continuous on [0, infinity) but is not

the square root function f is continuous on [0, infinity) . It is continuous at 0.
 
Imagine that this square root function was:

f(x) = 1/sqrt(x)

not only is it not continuous, but is undefined at zero.

Hope that helps,
Matt
 
You're looking at two claims:
(1) f is continuous on [0,?), and
(2) f is discontinuous at 0.
By definition, a function is continuous on an interval exactly when it is continuous at every point of the interval. So (1) and (2) cannot both be true within the same context.

But continuity depends very much on context.

The Calc 1 definition of continuity is that a function f is continuous at a point a if,
lim_{x->a} f(x) = f(a)

The kicker is the variable "x." What kind of values can it assume? Is x allowed to be positive? Negative? Complex? Vector-valued? Bunny rabbit valued?

Usually we require x to be in the domain of f, but it's sometimes useful to allow x to any element of some space containing the domain of f. This may offer an explanation. For if we regard the domain of f as embedded in the real numbers, then f is, loosely, a function from the real numbers to the real numbers. In this context f(x) is undefined for negative x. lim_{x->a} f(x) consequently does not exist, so f is discontinuous at 0.

Briefly, if R designates the real numbers and C the complex numbers, then
f=?x is continuous from the real interval [0,?) to R, and is continuous at 0 in this context.
f=?x is not continuous from R to R, and is not continuous at 0 in this context.
f=?x is continuous from R to C, and is continuous at 0 in this context.
f=?x is continuous from C to C and is continuous at 0 in this context.
(Strictly, we should say that f=?x can be defined in such a way that it is continuous from C to C.)
 
the square root function f is continuous on [0, infinity) . It is continuous at 0.
 
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