if ( f ) is continuous, ( f(1)=0 ), and ( f(1)=8 ), evaluate\n\n lim _{x \rightarrow 0} \frac{f(1+2 x)+f(1+6…

if ( f ) is continuous, ( f(1)=0 ), and ( f(1)=8 ), evaluate\n\n lim _{x \rightarrow 0} \frac{f(1+2 x)+f(1+6 x)}{x}.

if ( f ) is continuous, ( f(1)=0 ), and ( f(1)=8 ), evaluate\n\n lim _{x \rightarrow 0} \frac{f(1+2 x)+f(1+6 x)}{x}.

Answer

Explanation:

Step1: Use the definition of derivative

The definition of the derivative is (f^{\prime}(a)=\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}). For (\lim_{x\rightarrow0}\frac{f(1 + 2x)}{x}), let (h = 2x). As (x\rightarrow0), (h\rightarrow0). Then (\lim_{x\rightarrow0}\frac{f(1 + 2x)}{x}=2\lim_{h\rightarrow0}\frac{f(1+h)-f(1)}{h}). Since (f(1) = 0), this is (2f^{\prime}(1)). For (\lim_{x\rightarrow0}\frac{f(1 + 6x)}{x}), let (h = 6x). As (x\rightarrow0), (h\rightarrow0). Then (\lim_{x\rightarrow0}\frac{f(1 + 6x)}{x}=6\lim_{h\rightarrow0}\frac{f(1+h)-f(1)}{h}). Since (f(1) = 0), this is (6f^{\prime}(1)).

Step2: Calculate the limit

(\lim_{x\rightarrow0}\frac{f(1 + 2x)+f(1 + 6x)}{x}=\lim_{x\rightarrow0}\frac{f(1 + 2x)}{x}+\lim_{x\rightarrow0}\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)