Question

The value of f (0) so that the function 

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

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Answer 1

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

 For  \[f\left( x \right)\]  to be continuous at x = 0, we must have

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]

\[\Rightarrow f\left( 0 \right) = \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \frac{2 - \left( 256 - 7x \right)^\frac{1}{8}}{\left( 5x + 32 \right)^\frac{1}{5} - 2}\]
\[ \Rightarrow f\left( 0 \right) = \lim_{x \to 0} \frac{{256}^\frac{1}{8} - \left( 256 - 7x \right)^\frac{1}{8}}{\left( 5x + 32 \right)^\frac{1}{5} - {32}^\frac{1}{5}}\]
\[ = - \lim_{x \to 0} \frac{\frac{\left[ \left( 256 - 7x \right)^\frac{1}{8} - {256}^\frac{1}{8} \right]}{x}}{\frac{\left[ \left( 5x + 32 \right)^\frac{1}{5} - {32}^\frac{1}{5} \right]}{x}}\]
\[ = \frac{- 7}{5} \lim_{x \to 0} \frac{\frac{\left[ \left( 256 - 7x \right)^\frac{1}{8} - {256}^\frac{1}{8} \right]}{7x}}{\frac{\left[ \left( 5x + 32 \right)^\frac{1}{5} - {32}^\frac{1}{5} \right]}{5x}}\]
\[ = \frac{7}{5} \lim_{x \to 0} \frac{\frac{\left[ \left( 256 - 7x \right)^\frac{1}{8} - {256}^\frac{1}{8} \right]}{\left( 256 - 7x \right) - 256}}{\frac{\left[ \left( 5x + 32 \right)^\frac{1}{5} - {32}^\frac{1}{5} \right]}{5x + 32 - 32}}\]
\[ = \frac{7}{5} \times \frac{\frac{1}{8} \times \left( 256 \right)^{- \frac{7}{8}}}{\frac{1}{5} \times \left( 32 \right)^\frac{- 4}{5}}\]
\[ = \frac{7}{5} \times \frac{\frac{1}{8} \times 2^4}{\frac{1}{5} \times 2^7}\]
\[ = \frac{7}{64}\]