Question

If the function $f\left(x\right)=\frac{\log (1+ax)-\log (1-bx)}{x}$ for $x\ne 0$ is continuous at $x=0$ then $f\left(0\right)=$

• A. $a-b$
• B. $a+b$
• C. $\log a+\log b$
• D. $\log a-\log b$

Answer 1

The correct option is B $a+b$

Since, on applying the limit, the function is of $\frac{0}{0}$ form,
we apply the L-Hospital's Rule and differentiate both the numerator and the denominator individually.
Thus, the question transforms to
$\underset{x\to 0}{lim}\frac{a}{1+ax}-\frac{-b}{1-bx}$

Now, applying the limit we get the value as $a-(-b)=a+b$
Thus for the function to be continuous at $x=0$
$f\left(0\right)=a+b$