Question

what is \\(\lim_{h \to 0} \frac{8\left(\frac{1}{2} + h\
ight)^8 - 8\left(\frac{1}{2}\
ight)^8}{h}\\)?

Explanation:

Step1: Recall the definition of derivative

The limit \(\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\) is the definition of the derivative of the function \(y = f(x)\) at \(x\), that is \(f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\).

In the given limit \(\lim_{h
ightarrow0}\frac{8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}}{h}\), we can let \(f(x)=8x^{8}\) and \(x = \frac{1}{2}\).

Step2: Find the derivative of \(f(x)\)

First, we find the derivative of the function \(f(x)=8x^{8}\) using the power rule. The power rule states that if \(y = ax^{n}\), then \(y^{\prime}=nax^{n - 1}\).

For \(f(x)=8x^{8}\), where \(a = 8\) and \(n = 8\), by the power rule, \(f^{\prime}(x)=8\times8x^{8 - 1}=64x^{7}\).

Step3: Evaluate the derivative at \(x=\frac{1}{2}\)

Now we need to find the value of \(f^{\prime}(x)\) at \(x=\frac{1}{2}\). Substitute \(x = \frac{1}{2}\) into \(f^{\prime}(x)=64x^{7}\):

\(f^{\prime}(\frac{1}{2})=64\times(\frac{1}{2})^{7}\)

We know that \((\frac{1}{2})^{7}=\frac{1}{2^{7}}=\frac{1}{128}\), so:

\(f^{\prime}(\frac{1}{2})=64\times\frac{1}{128}\)

Simplify the right - hand side: \(64\times\frac{1}{128}=\frac{64}{128}=\frac{1}{2}\)

Answer:

\(\frac{1}{2}\)