Question

what is \\( \lim _ { h \
ightarrow 0 } \frac { 8 ( \frac { 1 } { 2 } + h ) ^ { 8 } - 8 ( \frac { 1 } { 2 } ) ^ { 8 } } { h } \\)?

Explanation:

Step1: Expand the expressions

First, expand \(8(\frac{1}{2}+h)^{8}\) and \(8(\frac{1}{2})^{8}\) using the binomial theorem \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\).
For \((\frac{1}{2}+h)^{8}=\sum_{k = 0}^{8}\binom{8}{k}(\frac{1}{2})^{8 - k}h^{k}=(\frac{1}{2})^{8}+8(\frac{1}{2})^{7}h+\cdots+h^{8}\).
Then \(8(\frac{1}{2}+h)^{8}=8[(\frac{1}{2})^{8}+8(\frac{1}{2})^{7}h+\cdots+h^{8}]=8(\frac{1}{2})^{8}+64(\frac{1}{2})^{7}h+\cdots+8h^{8}\).
And \(8(\frac{1}{2})^{8}\) is just a constant.
So the numerator \(8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}=64(\frac{1}{2})^{7}h+\cdots+8h^{8}\).

Step2: Simplify the limit expression

The limit \(\lim_{h
ightarrow0}\frac{8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}}{h}=\lim_{h
ightarrow0}\frac{64(\frac{1}{2})^{7}h+\cdots+8h^{8}}{h}\).
Factor out an \(h\) from the numerator: \(\lim_{h
ightarrow0}\frac{h[64(\frac{1}{2})^{7}+\cdots+8h^{7}]}{h}\).
Cancel out the \(h\) terms (since \(h
eq0\) when taking the limit as \(h
ightarrow0\) but \(h
eq0\) in the expression \(\frac{8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}}{h}\)), we get \(\lim_{h
ightarrow0}[64(\frac{1}{2})^{7}+\cdots+8h^{7}]\).

Step3: Evaluate the limit

As \(h
ightarrow0\), the terms with \(h\) (i.e., \(8h^{7},\cdots\)) go to \(0\).
We know that \(64(\frac{1}{2})^{7}=\frac{64}{128}=\frac{1}{2}\).

Answer:

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