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A continuous function is one where you can draw the graph of the function without lifting your pencil. This means there are no breaks, jumps, or holes in the graph. Mathematically, a function is continuous at a point if the limit as the function approaches that point equals the function's value at that point.
To check if a function is continuous at a specific point, say at \(x = a\), we often perform these steps:

• Calculate \(f(a)\), the function's value at \(a\).
• Determine the limit of the function as \(x\) approaches \(a\), or \(\lim_{{x \to a}} f(x)\).
• Ensure that both \(\lim_{{x \to a}} f(x)\) and \(f(a)\) exist and are equal.

In the given problem, we simplified the function \(f(x) = \frac{x^2 - 5x + 6}{x - 2}\) to \(f(x) = x - 3\) for \(x eq 2\). The limit as \(x\) approaches \(2\) is \(-1\), which is the same as \(f(2) = -1\). Therefore, the function is continuous at \(x = 2\).

The derivative of a function gives us the rate at which the function's value is changing at any given point. It's like finding the slope of the tangent line to the function's graph at a specific point. If a function has a derivative at a certain point, it means the function is smooth (no sharp corners or cusps) at that point.
For the problem at hand, we first simplify our function to \(f(x) = x - 3\), a straight line with a constant slope. The derivative, \(f'(x)\), of this linear function is \(1\) everywhere, indicating a constant, steady change without interruptions.
Because the derivative is a constant, it exists throughout the function's domain, including at \(x = 2\). This constant derivative shows that the function is differentiable at \(x = 2\). Differentiability implies continuity, but continuity alone doesn't imply differentiability. Thus, having a derivative confirms that any slight changes around \(x = 2\) remain smooth.

Limits help us approach very ‘close to’ a certain point on a function. They're crucial in analyzing the behavior of functions as they approach certain points. Limits can be thought of as what a function 'wants' to be as it gets infinitely close to a point.
In the function \(f(x) = \frac{x^2 - 5x + 6}{x - 2}\), the direct substitution of \(x = 2\) would cause a zero in the denominator, leading to an undefined expression. But by simplifying the function to \(f(x) = x - 3\) when \(x eq 2\), we find the limit as \(x\) approaches \(2\) is \(\lim_{{x \to 2}} (x - 3) = -1\).
This limit is a crucial part of verifying the function's continuity. By showing that the limit as \(x\) approaches \(2\) equals the function's value at \(x = 2\) (which is \(-1\)), we confirm that no jumps or gaps exist at that point. Limits ensure we can handle functions' behavior at points of potential trouble effectively.