91Ó°ÊÓ

We need to find the limit of the function as \(x\) approaches 2. If this limit exists and is equal to the function value at \(x=2\), then the function is continuous at \(x=2\). Otherwise, the function is not continuous at \(x=2\). The function is given by \(f(x) = \frac{x^{2}-5 x+6}{x-2}\). Factor the numerator: \(x^{2}-5 x+6 = (x-2)(x-3)\). Then, the function can be written as $$f(x) = \frac{(x-2)(x-3)}{x-2}$$ which simplifies to $$f(x) = \begin{cases} x-3 & \text{if } x \neq 2 \\ \text{undefined} & \text{if } x=2 \end{cases} $$ Now, we find the limit as \(x\) approaches 2: $$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x-3) = 2 - 3 = -1$$ As \(f(2)\) is undefined but the limit exists, the function is not continuous at \(x=2\).

Since the function is not continuous at \(x=2\), it is not differentiable at \(x=2\).