Question

The value of f (0), so that the function

\[f\left( x \right) = \frac{\left( 27 - 2x \right)^{1/3} - 3}{9 - 3 \left( 243 + 5x \right)^{1/5}}\left( x \neq 0 \right)\] is continuous, is given by 

पर्याय

• \[\frac{2}{3}\]

• 6

• 2

• 4

Answer 1

2 

For f(x) to be continuous at x = 0, we must have 

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow f\left( 0 \right) = \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \frac{\left( 27 - 2x \right)^\frac{1}{3} - 3}{9 - 3 \left( 243 + 5x \right)^\frac{1}{5}}\]
\[ \Rightarrow f\left( 0 \right) = \lim_{x \to 0} \frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{3\left( {243}^\frac{1}{5} - \left( 243 + 5x \right)^\frac{1}{5} \right)}\]
\[ = \frac{1}{3} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{x}}{\frac{\left( {243}^\frac{1}{5} - \left( 243 + 5x \right)^\frac{1}{5} \right)}{x}}\]
\[ = \frac{- 1}{3} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{x}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{x}}\]
\[ = \frac{2}{15} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{- 2x}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{5x}}\]
\[ = \frac{2}{15} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{27 - 2x - 27}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{243 + 5x - 243}}\]
\[ = \frac{2}{15} \times \frac{\frac{1}{3} \times {27}^\frac{- 2}{3}}{\frac{1}{5} \times {243}^\frac{- 4}{5}}\]
\[ = \frac{2}{15} \times \frac{\frac{1}{3} \times \frac{1}{{27}^\frac{2}{3}}}{\frac{1}{5} \times \frac{1}{{243}^\frac{4}{5}}}\]
\[ = 2\]