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# In the function $f\left( x \right) = \dfrac{{\log \left( {1 + ax} \right) - \log \left( {1 - bx} \right)}}{x},x \ne 0$ is continuous at $x = 0$, then $f\left( 0 \right) =$A. $\log a - \log b$B. $a + b$C. $\log a + \log b$D. $a - b$ Verified
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Hint: Use property of limits i.e. $\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + x} \right)}}{x} = 1$

Given function $f\left( x \right) = \dfrac{{\log \left( {1 + ax} \right) - \log \left( {1 - bx} \right)}}{x}$
For $f\left( x \right)$to be continuous, we must have $f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} f\left( x \right)$
Put the value of$f\left( x \right)$, we get
$f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + ax) - \log (1 - bx)}}{x}$
Multiply and divide the equation with $a$ and $- b$, we get
$f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{a\log (1 + ax)}}{{ax}} - \dfrac{{\left( { - b} \right)\log (1 - bx)}}{{ - bx}}$
We know that, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + x} \right)}}{x} = 1$
$\therefore f\left( 0 \right) = a.1 + b.1 \\ f\left( 0 \right) = a + b \\$
Hence, the correct option is B.

Note: Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits.