Question

if ( f ) is continuous, ( f(1)=0 ), and ( f(1)=8 ), evaluate

lim _{x
ightarrow 0} \frac{f(1+2 x)+f(1+6 x)}{x}.

Explanation:

Step1: Use the definition of derivative

The definition of the derivative is \(f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}\).
For \(\lim_{x
ightarrow0}\frac{f(1 + 2x)}{x}\), let \(h = 2x\). As \(x
ightarrow0\), \(h
ightarrow0\). Then \(\lim_{x
ightarrow0}\frac{f(1 + 2x)}{x}=2\lim_{h
ightarrow0}\frac{f(1+h)-f(1)}{h}\). Since \(f(1) = 0\), this is \(2f^{\prime}(1)\).
For \(\lim_{x
ightarrow0}\frac{f(1 + 6x)}{x}\), let \(h = 6x\). As \(x
ightarrow0\), \(h
ightarrow0\). Then \(\lim_{x
ightarrow0}\frac{f(1 + 6x)}{x}=6\lim_{h
ightarrow0}\frac{f(1+h)-f(1)}{h}\). Since \(f(1) = 0\), this is \(6f^{\prime}(1)\).

Step2: Calculate the limit

\(\lim_{x
ightarrow0}\frac{f(1 + 2x)+f(1 + 6x)}{x}=\lim_{x
ightarrow0}\frac{f(1 + 2x)}{x}+\lim_{x
ightarrow0}\frac{f(1 + 6x)}{x}\)
Substitute the results from Step1: \(2f^{\prime}(1)+6f^{\prime}(1)\)
Given \(f^{\prime}(1)=8\), then \(2\times8 + 6\times8=(2 + 6)\times8\)
\(=8\times8\)

Answer:

\(64\)