If the function f(x) = \ array ll (1)/(x) _ e ((1 + (x)/(a))/(1 -...

Given that

$f\left(x\right) =$\left\{$$\begin{array}{ll}\dfrac{1}{x}\log_{e}\left(\dfrac{1 + \dfrac{x}{a}}{1 - \dfrac{x}{b}}\right), x 0\end{array}$$\right)

$f\left(x\right)$ is continuous at $x = 0$

$L . H . L = R . H . L = f\left(0\right)$

$LHL = \lim_{x \to {0}^{-}}\dfrac{\log \left(\dfrac{1 + \dfrac{x}{a}}{1 - \dfrac{x}{b}}\right)}{x}$

$= \lim_{x \to {0}^{-}}\dfrac{\log \left(1 + \dfrac{x}{a}\right) - \log \left(1 - \dfrac{x}{b}\right)}{x}$

$= \lim_{x \to {0}^{-}}\dfrac{\log \left(1 + \dfrac{x}{a}\right)}{\dfrac{x}{a}} \times \dfrac{1}{a} - \dfrac{\log \left(1 - \dfrac{x}{b}\right)}{- \dfrac{x}{b}} \times - \dfrac{1}{b} = \dfrac{1}{b} + \dfrac{1}{a}$

$RHL = \lim_{x \to {0}^{+}}\dfrac{\cos^{2}x - \sin^{2}x - 1}{\sqrt{{x}^{2} + 1} - 1} \times \dfrac{\sqrt{{x}^{2} + 1} + 1}{\sqrt{{x}^{2} + 1} + 1}$

$= \lim_{x \to {0}^{+}} - \dfrac{2\sin^{2}x}{{x}^{2}}\left(\sqrt{{x}^{2} + 1} + 1\right) = - 4$

So, $\dfrac{1}{b} + \dfrac{1}{a} = k = - 4 .$.

$k = - 4 ⇒ \dfrac{4}{k} = - 1 .$

$\dfrac{1}{a} + \dfrac{1}{b} = - 4 .$

$∴ \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{4}{k} = - 5 .$