$f(x)=\frac{2-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-2}, x \neq 0$ is continuous everywhere, is given by
(a) $-1$
(b) 1
(c) 26
(d) none of these
(d) none of these
Given: $f(x)=\frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$
For $f(x)$ to be continuous at $x=0$, we must have
$\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$
$\Rightarrow f(0)=\lim _{x \rightarrow 0} \frac{256^{\frac{1}{8}}-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-32 \frac{1}{5}}$
$=\frac{7}{5} \times \frac{\frac{1}{8} \times(256)^{-\frac{7}{8}}}{\frac{1}{5} \times(32)^{\frac{-4}{5}}}$
$=\frac{7}{5} \times \frac{\frac{1}{8} \times 2^{4}}{\frac{1}{5} \times 2^{7}}$
$=\frac{7}{64}$