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**Hint**: Use the conditions of limit to test continuity and then try to convert the given function into the standard form of $ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log (1+x)}{x}=1 $ . \[\]

**:**

__Complete step-by-step answer__We are going to use limits to test continuity at the point where the function is not defined. If a function $ f\left( x \right) $ is continuous at any point $ x=a $ then function $ f\left( x \right) $ must satisfy 3 following conditions \[\]

1. The function must be defined at $ x=a $ that means $ f\left( a \right) $ has to be a real number. \[\]

2. The limit of the function must exist at the point that means both the left hand limit and the right hand limit must exist and have to be equal to some real number. In symbols for $ \mathop {\lim }\limits_{x \to {a^ - }} \left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) $ \[\]

3. The limiting value of the function and the exact value of value at the point that means at $ x=a $ have to be equal $ \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) $ . \[\]

As asked in the question $ f\left( x \right) $ is not defined at $ x=0 $ but has to be continuous at $ x=0 $ . If $ f\left( x \right) $ is continuous at $ x=0 $ then from second condition we get that the limit exists and from third condition we get $ \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=f\left( 0 \right) $ \[\]

\[\begin{align}

& \therefore \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=f\left( 0 \right) \\

& \Rightarrow f\left( 0 \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+ax \right)-\log \left( 1-bx \right)}{x} \\

& =a\underset{ax\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+ax \right)}{ax}+b\underset{-bx\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+\left( -bx \right) \right)}{-bx}\left( \because x\to 0,ax\to \text{ }and\text{ }-bx\to 0 \right) \\

& =a\cdot 1+b\cdot 1\text{ }=a+b\text{ }\left( \text{Using }\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+x \right)}{x}=1 \right) \\

\end{align}\]

So the correct choice is (A) $ a+b $ .

**Note**: The question tests your concept of limits as a precondition of continuity that means we have to use limits to test continuity. Here the calculation limits requires us to use the formula and we have to bring it to the standard so that we can use it.

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