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)?

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?

To determine if the function f(x) = 1/sqrt(x-5) is continuous at x=2, we need to check the three conditions for continuity:

1. The function should be defined at x=2.
2. The limit of the function as x approaches 2 should exist.
3. The value of the function at x=2, f(2), should be equal to the limit as x approaches 2, lim x->2 f(x).

Let's go through each condition:

1. The function f(x) = 1/sqrt(x-5) is defined for all values of x except x=5. Since x=2 is not equal to x=5, the function is defined at x=2.

2. To find the limit as x approaches 2, we can substitute x=2 into the function: lim x->2 f(x) = lim x->2 1/sqrt(x-5) = 1/sqrt(2-5) = 1/sqrt(-3). However, the square root of a negative number is not defined in the real number system, so this limit does not exist.

3. Since the limit as x approaches 2 does not exist, there is no value of f(2) to compare it to. Therefore, we cannot determine if f(2) is equal to the limit.

Based on these three conditions, we can conclude that f(x) = 1/sqrt(x-5) is not continuous at x=2.

To determine if a function is continuous at a specific point, we need to check three conditions:

1) The function must be defined at that point.
2) The limit of the function as x approaches that point must exist.
3) The value of the function at that point must be equal to the limit at that point.

Let's apply these conditions to the function f(x) = 1/√(x-5) at x = 2:

1) The function must be defined at x = 2. In this case, the function is defined for all x except 5. Since 2 is not equal to 5, the function is defined at x = 2.

2) The limit of the function as x approaches 2 must exist. We can find this limit by evaluating the limit of the function as x approaches 2:

lim (x->2) 1/√(x-5)

To evaluate this limit, we can substitute x = 2 into the function:

1/√(2-5) = 1/√(-3)

Since the square root of a negative number is not defined in the real number system, the limit does not exist. Therefore, the function is not continuous at x = 2.

3) Since the limit in condition 2 does not exist, the third condition is not applicable.

In summary, the function f(x) = 1/√(x-5) is not continuous at x = 2.