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The limit definition of a derivative is essential in calculus. It helps us find the slope of the tangent line to a curve at any given point. For a function, say \(y = f(x)\), the derivative at a specific point \(x = a\) is defined by:\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]This formula shows how the slope of the function is determined by the subtle changes around the point \(a\) as \(h\), an infinitesimally small number, approaches zero. Here, the expression \(f(a+h) - f(a)\) represents the rise (change in y) and \(h\) represents the run (change in x). This limit picks out the exact rate of change of the function at the point \(x = a\).
In our specific exercise, using the limit definition, we matched the form to identify that our function \(f(x) = \sqrt{x}\) at the point \(a = 2\).

Rationalizing the numerator is a technique often used to simplify expressions, especially when limits involving radicals are involved. In the given example, the expression \(\frac{\sqrt{2+h} - \sqrt{2}}{h}\) needed simplification.
To do this, we multiply the expression by the conjugate of the numerator. The conjugate of \(\sqrt{2+h} - \sqrt{2}\) is \(\sqrt{2+h} + \sqrt{2}\). This allows us to eliminate the radical in the numerator through the difference of squares:

• \(\left(\sqrt{2+h}\right)^2 - \left(\sqrt{2}\right)^2 = (2+h) - 2 = h\)

Subsequently, the numerator becomes \(h\), which now cancels with the \(h\) in the denominator, leaving us with an expression that can be easily evaluated as \(h\) approaches zero: \(\frac{1}{\sqrt{2+h} + \sqrt{2}}\).

The slope of a tangent line to a curve at a given point is an important concept in calculus as it represents the instantaneous rate of change of the function at that point.
Once we have simplified the derivative expression, evaluating the limit gives us the slope.In our example, after simplification, we obtained:\[f'(2) = \frac{1}{\sqrt{2+h} + \sqrt{2}}\]And by evaluating this limit as \(h\) approaches zero, we find:
\[f'(2) = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}\]This result tells us about the steepness of the curve \(y = \sqrt{x}\) at the point \(x = 2\). The value \(\frac{\sqrt{2}}{4}\) is the exact slope of the tangent line at that point, reflecting how sharply the function is changing there.

Function evaluation is a routine step in calculus when you're supposed to replace the variable with a specific value and calculate the result. In this context, it's directly tied to finding the derivatives and understanding function behaviors.
Evaluating a function presence shows through in different steps, such as in recognizing which function form you're dealing with or after finding a derivative.
In our exercise, after noting that the function is \(f(x) = \sqrt{x}\), we found \(f(2) = \sqrt{2}\). These evaluations give points on the function that are used to understand its slope at that specific point by calculating its derivative, as done through the limit definition. Function evaluation helps verify work and connect calculations back to the original problem definition.