Question

If $f\left(x\right)=\frac{2-{(256-7x)}^{1/8}}{{(5x+32)}^{1/5}-2}(x\ne 2),$ then for $f$ to be continuous everywhere,$f\left(0\right)$ is

• A. $\frac{7}{64}$
• B. $\frac{7}{68}$
• C. $\frac{5}{64}$
• D. $\frac{5}{68}$

Answer 1

The correct option is A $\frac{7}{64}$

$f\left(x\right)=\frac{2-{(256-7x)}^{1/8}}{{(5x+32)}^{1/5}-2}$

For $f\left(x\right)$ to be continuous everywhere

$\underset{x\to 0}{lim}f\left(x\right)=$finite

$\Rightarrow \underset{x\to 0}{lim}\frac{2-{(256-7x)}^{1/8}}{{(5x+32)}^{1/5}-2}\frac{0}{0}$ form

$\underset{x\to 0}{lim}\frac{0-\frac{1}{8}{(256-7x)}^{-7/8}\times (-7)}{{\frac{1}{5}(5x+32)}^{-4/5}\left(5\right)-0}$

$=\frac{7\times 5}{8\times 5}\frac{{\left(256\right)}^{-7/8}}{{\left(32\right)}^{-4/5}}=\frac{7\times 2^4}{8\times 2^7}$

$=\frac{7}{64}$