Final answer:
The function f(x) = 1/(x-2) is continuous on the interval [1,3] excluding x=2. Continuous probability density functions define probabilities as areas under the curve, and the probability of a specific point in a continuous distribution is 0.
Step-by-step explanation:
The question encompasses concepts from the subject of Mathematics, specifically from the unit on continuity and probability functions within calculus and probability theory. In the given scenario, without the graph of the function f(x), we cannot directly determine the limit as x approaches 1 or the value of f(1). However, a function is continuous at a point if the limit as x approaches that point equals the function's value at that point, and the function is defined there. The function f(x)=1/(x-2) has a discontinuity at x=2, since the denominator becomes zero and the function is undefined. However, on the interval [1,3] excluding x=2, it is continuous because there are no other points where the function is not defined or where the limit does not equal the function value.
Continuous probability density functions are used to define probabilities as areas under the curve of the function over an interval on the x-axis. Therefore, the probability of a specific point, like P(x=7) in a continuous distribution, is always 0 since a point has no area.