Question

The value of f (0) so that the function
$f\left(x\right)=\frac{2-{\left(256-7x\right)}^{1/8}}{{\left(5x+32\right)}^{1/5}-2},$ x ≠ 0 is continuous everywhere, is given by
(a) −1
(b) 1
(c) 26
(d) none of these

Answer 1

(d) none of these

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

For $f\left(x\right)$ to be continuous at x = 0, we must have
$\underset{\mathrm{x}\to 0}{\lim }f\left(x\right)=f\left(0\right)$
$$\begin{gathered} \Rightarrow f\left(0\right)=\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{2-{\left(256-7x\right)}^{{\displaystyle \frac{1}{8}}}}{{\left(5x+32\right)}^{{\displaystyle \frac{1}{5}}}-2} \\ \Rightarrow f\left(0\right)=\underset{x\to 0}{\lim }\frac{{256}^{{\displaystyle \frac{1}{8}}}-{\left(256-7x\right)}^{{\displaystyle \frac{1}{8}}}}{{\left(5x+32\right)}^{{\displaystyle \frac{1}{5}}}-{32}^{{\displaystyle \frac{1}{5}}}} \\ =-\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\left[{\left(256-7x\right)}^{{\displaystyle \frac{1}{8}}}-{256}^{\frac{1}{8}}\right]}{x}}}{{\displaystyle \frac{\left[{\left(5x+32\right)}^{{\displaystyle \frac{1}{5}}}-{32}^{{\displaystyle \frac{1}{5}}}\right]}{x}}} \\ =\frac{-7}{5}\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\left[{\left(256-7x\right)}^{\frac{1}{8}}-{256}^{\frac{1}{8}}\right]}{7x}}}{{\displaystyle \frac{\left[{\left(5x+32\right)}^{{\displaystyle \frac{1}{5}}}-{32}^{{\displaystyle \frac{1}{5}}}\right]}{5x}}} \\ =\frac{7}{5}\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\left[{\left(256-7x\right)}^{\frac{1}{8}}-{256}^{\frac{1}{8}}\right]}{\left(256-7x\right)-256}}}{{\displaystyle \frac{\left[{\left(5x+32\right)}^{{\displaystyle \frac{1}{5}}}-{32}^{{\displaystyle \frac{1}{5}}}\right]}{5x+32-32}}} \\ =\frac{7}{5}\times \frac{{\displaystyle \frac{1}{8}}\times {\left(256\right)}^{-{\displaystyle \frac{7}{8}}}}{{\displaystyle \frac{1}{5}}\times {\left(32\right)}^{{\displaystyle \frac{-4}{5}}}} \\ =\frac{7}{5}\times \frac{{\displaystyle \frac{1}{8}}\times 2^4}{{\displaystyle \frac{1}{5}}\times 2^7} \\ =\frac{7}{64} \end{gathered}$$