Question

If $f\left(0\right)=1$ & $f^{\prime }\left(0\right)=9$, then ${lim}_{x\to 0}\frac{f\left(x\right)-f\left(\frac{x}{2}\right)+f\left(\frac{x}{2^2}\right)-f\left(\frac{x}{2^3}\right)}{f\left(x\right)-1}$ is equal to-

• A. $\frac{2}{7}$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{7}{9}$

Answer 1

The correct option is C $\frac{5}{8}$

$f^{\prime }\left(0\right)=\underset{n\to 0}{lim}\frac{f\left(n\right)-f\left(0\right)}{n}$

$\therefore 9=\underset{x\to 0}{lim}\frac{f\left(x\right)-1}{x}$

$\therefore \underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(x/2\right)+f\left(\frac{x}{4}\right)-f\left(x/8\right)}{f\left(x\right)-1}$

$=\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)-\frac{1}{2}{f}^{\prime }\left(x/2\right)+\frac{1}{4}{f}^{\prime }\left(x/4\right)-\frac{1}{8}{f}^{\prime }\left(x/8\right)}{f^{\prime }\left(x\right)}$

$=\frac{9-\frac{9}{2}+\frac{9}{4}-\frac{9}{8}}{9}$

$=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}$

$=\frac{8-4+2-1}{8}$

$=\frac{5}{8}$