91Ó°ÊÓ

The limit definition of the derivative is a fundamental concept in calculus. It helps us understand how functions change at any given point. For a function \(f(x)\), the derivative at a point \(x = a\) is defined using limits as:\[ f^{\prime}(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]This formula tells us the rate at which the function \(f(x)\) is changing at \(x = a\).

To apply this to our example, we want to find the derivative of \(f(x) = \frac{1}{x+1}\) at \(x=1\), which is \(f^{\prime}(1)\). By substituting \(1\) into the formula, it becomes:\[ f^{\prime}(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \]

This setup is essential for finding the exact slope of the function at a specific point.

Simplifying algebraic expressions involves making expressions easier to work with. This often means eliminating complex fractions or solving for a common denominator.

In our problem, we have the expression \(\frac{1}{2+h} - \frac{1}{2}\) within the limit formula. To simplify it, we find a common denominator, which in this case is \((2+h)*2\). Thus, it becomes:\[\frac{1}{2+h} - \frac{1}{2} = \frac{2 - (2+h)}{(2+h) \cdot 2} = \frac{-h}{2(2+h)}\]

This simplification allows us to cancel out \(h\) in the numerator and denominator, greatly simplifying the limit calculation in the next step.

Evaluating limits is a critical step in calculus, as it helps us determine the value a function approaches as the input approaches a certain point. After simplifying the expression in our derivative problem, we look at:\[f^{\prime}(1) = \lim_{h \to 0} \frac{-h}{2h(2+h)}\]

Here, we can cancel out \(h\) in the numerator and denominator because \(h\) is approaching 0 but never exactly 0 in the limit process. This gives:\[f^{\prime}(1) = \lim_{h \to 0} \frac{-1}{2(2+h)}\]

Now, by substituting \(h=0\), we find:\[f^{\prime}(1) = \frac{-1}{4}\]

This tells us that the slope of the function \(f(x) = \frac{1}{x+1}\) at \(x=1\) is \(-\frac{1}{4}\).