Question

If the function f (x) defined by  \[f\left( x \right) = \begin{cases}\frac{\log \left( 1 + 3x \right) - \log \left( 1 - 2x \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then k =

 

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Answer 1

 Given: 

\[f\left( x \right) = \binom{\frac{\log \left( 1 + 3x \right) - \log \left( 1 - 2x \right)}{x}, x \neq 0}{k, x = 0}\]

If f(x) is continuous at x = 0, then​

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

\[\Rightarrow \lim_{x \to 0} \left( \frac{\log\left( 1 + 3x \right) - \log\left( 1 - 2x \right)}{x} \right) = k\]

\[\Rightarrow \lim_{x \to 0} \left( \frac{3 \log \left( 1 + 3x \right)}{3x} - \frac{2 \log \left( 1 - 2x \right)}{2x} \right) = k\]
\[ \Rightarrow 3 \lim_{x \to 0} \left( \frac{\log \left( 1 + 3x \right)}{3x} \right) - 2 \lim_{x \to 0} \left( \frac{\log \left( 1 - 2x \right)}{2x} \right) = k\]
\[ \Rightarrow 3 \lim_{x \to 0} \left( \frac{\log \left( 1 + 3x \right)}{3x} \right) + 2 \lim_{x \to 0} \left( \frac{\log \left( 1 - 2x \right)}{- 2x} \right) = k\]
\[ \Rightarrow 3 \times 1 + 2 \times 1 = k \left[ \because \lim_{x \to 0} \frac{\log \left( 1 + x \right)}{x} = 1 \right]\]
\[ \Rightarrow k = 3 + 2 = 5\]