6) What are the three conditions that determine the continuity of a function? \( \begin{array}{llll}\text { ()) } f(x)=\frac{x^{3}-1}{x-1} & x \neq 1 & \text { b) ) } f(x)=\frac{x^{3}-1}{x-1} \quad x \neq 1 \\ \text { 2, } x=1 & 3, & x=1\end{array} \)

To decide whether a function is continuous at a point, there are three necessary conditions: 1. The function must be defined at the point. In other words, f(c) exists. 2. The limit of the function as x approaches the point must exist. That is, limₓ→c f(x) exists. 3. The value of the function at the point must equal the limit. That is, limₓ→c f(x) = f(c). Let’s apply these conditions to the function   f(x) = (x³ – 1)/(x – 1). Notice that (x³ – 1) can be factored using the difference of cubes:   x³ – 1 = (x – 1)(x² + x + 1). Thus, for x ≠ 1, we have   f(x) = (x – 1)(x² + x + 1)/(x – 1) = x² + x + 1. Now consider the following two cases regarding the definition of f(x) at x = 1: Case a) f is defined only for x ≠ 1. - Condition 1 fails at x = 1, because f(1) is not defined. - Even though limₓ→1 f(x) exists and equals 1² + 1 + 1 = 3, the function is not continuous at x = 1 because we cannot compare the limit to f(1). Case b) The function is defined piecewise by:   f(x) = (x³ – 1)/(x – 1) for x ≠ 1, and f(1) is defined to be 2. - Condition 1 is met because f(1) is defined. - Condition 2 is met because limₓ→1 f(x) = 1² + 1 + 1 = 3 exists. - However, Condition 3 fails since f(1) = 2 does not equal the limit 3. Therefore, the function is not continuous at x = 1. If instead we define f(1) = 3, then all three conditions would be satisfied at x = 1, and the function would be continuous there. In summary, the three conditions for continuity at a point c are:   (1) f(c) is defined.   (2) limₓ→c f(x) exists.   (3) limₓ→c f(x) = f(c). Using these, one can check the continuity of any function at a given point.