Question - Evaluating the Limit of a Function as It Approaches a Point

The expression given is the definition of the derivative of \(\sqrt{x}\) evaluated at \(x=8\), which can be calculated as follows:

\(\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\) at \(x=8\)

Let's apply the definition of the derivative for \(f(x) = \sqrt{x}\) at \(x = 8\):

\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

\(f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\)

To evaluate the limit, multiply the numerator and the denominator by the conjugate of the numerator:

\(= \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}\)

\(= \lim_{h \to 0} \frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}\)

\(= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}\)

\(= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\)

Now plug in \(x = 8\):

\(f'(8) = \lim_{h \to 0} \frac{1}{\sqrt{8+h} + \sqrt{8}}\)

\(= \frac{1}{\sqrt{8} + \sqrt{8}}\)

\(= \frac{1}{2\sqrt{8}}\)

\(= \frac{1}{4\sqrt{2}}\)

\(= \frac{\sqrt{2}}{8}\)

Therefore, the limit and the derivative of \(\sqrt{x}\) at \(x = 8\) is \(\frac{\sqrt{2}}{8}\).