Question

Let  \[f\left( x \right) = \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x}\] x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

 

Answer 1

Given: 

\[f\left( x \right) = \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x}, x \neq 0\]

If f(x) is continuous at x = 0, then

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

\[\Rightarrow \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x} \right) = f\left( 0 \right)\]

\[\Rightarrow \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right)}{\frac{ax}{a}} - \frac{\log\left( 1 - \frac{x}{b} \right)}{\frac{bx}{b}} \right) = f\left( 0 \right)\]

\[\Rightarrow \frac{1}{a} \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right)}{\frac{x}{a}} \right) - \left( - \frac{1}{b} \right) \lim_{x \to 0} \left( \frac{\log\left( 1 - \frac{x}{b} \right)}{\frac{- x}{b}} \right) = f\left( 0 \right)\]

\[\Rightarrow \frac{1}{a} \times 1 - \left( - \frac{1}{b} \right) \times 1 = f\left( 0 \right) \left[ \text{ Using } : \lim_{x \to 0} \frac{\log\left( 1 + x \right)}{x} = 1 \right]\]
\[ \Rightarrow \frac{1}{a} + \frac{1}{b} = f\left( 0 \right)\]
\[ \Rightarrow \frac{a + b}{ab} = f\left( 0 \right)\]