Solve this

Question:

$f(x)=\frac{2-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-2}, x \neq 0$ is continuous everywhere, is given by

(a) $-1$

(b) 1

(c) 26

(d) none of these

Solution:

(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

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$

$\Rightarrow f(0)=\lim _{x \rightarrow 0} \frac{256^{\frac{1}{8}}-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-32 \frac{1}{5}}$

$=\frac{7}{5} \times \frac{\frac{1}{8} \times(256)^{-\frac{7}{8}}}{\frac{1}{5} \times(32)^{\frac{-4}{5}}}$

$=\frac{7}{5} \times \frac{\frac{1}{8} \times 2^{4}}{\frac{1}{5} \times 2^{7}}$

$=\frac{7}{64}$

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