Question

Is f(x)=1/sqrt(x-5) continious at x=2?

I know it is discontinuous at x=5 but then I got confused with the three rules. Because function is continuous if f(a) a is in the domain, limit exits, and also lim x->a f(x) = f(a)?

Answer 1

The rules for the continuity of f(x) at a given point a are:
1. If a is an interior point,
lim x->a f(x) = f(a)
2. If a is a left end-point:
lim x->a+ f(x) = f(a)
3. If a is a right end-point:
lim x->a- f(x) = f(a)

For the above conditions to be satisfied,
1. f(a) must exist, i.e. a is in the domain of f(x).
2. The specified limits must exist, i.e.
for interior points,
lim x->a- f(x) must exist and
lim x->a+ f(x) must exist and they are both equal to f(a).
For exterior points, the one-side limit must exist and must be equal to f(a).

In case where the limits from each side are equal, but f(a) does not exist, then f(x) is discontinuous at a, and it is called a removable discontinuity.


For f(x)=1/sqrt(x-5) is x=2 in the domain?