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

Explanation:

Step1: Expand the expression

First, expand (8(\frac{1}{2}+h)^{8}) using the binomial theorem ((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}). Here (a=\frac{1}{2}), (b = h), and (n = 8).

(8(\frac{1}{2}+h)^{8}=8\sum_{k=0}^{8}\binom{8}{k}(\frac{1}{2})^{8 - k}h^{k}=8\left[\binom{8}{0}(\frac{1}{2})^{8}+\binom{8}{1}(\frac{1}{2})^{7}h+\binom{8}{2}(\frac{1}{2})^{6}h^{2}+\cdots+\binom{8}{8}h^{8}\right])

(=8\times\frac{1}{256}+8\times8\times\frac{1}{128}h + 8\times28\times\frac{1}{64}h^{2}+\cdots+8h^{8}=\frac{1}{32}+4h + 28h^{2}+\cdots+8h^{8})

And (8(\frac{1}{2})^{8}=\frac{1}{32})

So the numerator (8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}=4h + 28h^{2}+\cdots+8h^{8})

Step2: Simplify the limit

Now, (\lim_{h\rightarrow0}\frac{8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}}{h}=\lim_{h\rightarrow0}\frac{4h + 28h^{2}+\cdots+8h^{8}}{h})

Factor out (h) from the numerator: (\lim_{h\rightarrow0}\frac{h(4 + 28h+\cdots+8h^{7})}{h})

Cancel out the (h) terms ((h\neq0) when taking the limit as (h\rightarrow0) but (h\neq0) in the algebraic manipulation before taking the limit), we get (\lim_{h\rightarrow0}(4 + 28h+\cdots+8h^{7}))

Step3: Evaluate the limit

Substitute (h = 0) into the expression (4 + 28h+\cdots+8h^{7}), we have (4+28\times0+\cdots+8\times0^{7}=4)

Another way: We know that by the definition of the derivative (f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h})

Let (f(x)=8x^{8}), then (f^{\prime}(x)=64x^{7})

When (x=\frac{1}{2}), (f^{\prime}(\frac{1}{2})=64\times(\frac{1}{2})^{7}=4)

Answer:

(4)