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Limits are fundamental to understanding calculus and the concept of derivatives. When we evaluate a limit as a variable approaches a certain value, we are looking at the behavior of the function around that point. In the context of derivatives, limits help determine the instantaneous rate of change of a function at a specific point. This is crucial for problems involving rates and motion. For example, in our step-by-step solution, we evaluated the limit \[ \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \]by observing what happens as \(h\) gets infinitesimally small. The result provides the slope of the tangent line to the graph of the function at a given point. Therefore, understanding limits is key to grasping how changes affect the function's output as the input varies slightly.

The difference quotient is a major concept when dealing with derivatives. It measures the average rate of change over an interval and forms the basis of calculating a derivative. In simpler terms, the difference quotient for a function \( f(x) \) is given by:\[ \frac{f(x+h) - f(x)}{h} \]This expression calculates how much the function \(f(x)\) changes concerning a small increment \(h\). First, we test its behavior as \(h\) approaches zero. Doing so with the function \(f(x) = \frac{1}{x}\) at \(x = -2\), we substituted into the difference quotient, and through careful simplification steps, found it evaluates to a straightforward form. This calculated value represents the instantaneous slope of the curve at \(x = -2\), capturing precisely how the function changes at that point.

Function evaluation is a skill that involves substituting specific values into a function to determine its output. It is an essential part of solving calculus problems, especially when working with limits and derivatives. In our problem, we began by substituting \(x = -2\) into the function \(f(x) = \frac{1}{x}\) to find \(f(-2)\). This substitution transforms the function into a specific numerical result, which is necessary for forming the difference quotient:\[ f(-2) = -\frac{1}{2} \]Similarly, substituting \(x + h = -2 + h\) enables us to find \(f(-2 + h)\). Evaluating the function at specific points was the first step in simplifying the overall expression for the derivative. So, practice regularly evaluating functions to become proficient in understanding their outputs and behaviors.