For $$a, b > 0$$, let $$f(x) = \begin{cases} \frac{\tan((a+1)x) + b\tan x}{x}, & x < 0 \\ 3, & x = 0 \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b\sqrt{ax}\sqrt{x}}, & x > 0 \end{cases}$$ be a continuous function at $$x = 0$$. Then $$\frac{b}{a}$$ is equal to :

For continuity at $$x = 0$$, $$\text{LHL} = \text{RHL} = f(0) = 3$$.

• LHL: $$\lim_{x \to 0^-} \frac{\tan((a+1)x) + b\tan x}{x} = (a+1) + b = 3 \implies a + b = 2$$

• RHL: Rationalizing $$\lim_{x \to 0^+} \frac{\sqrt{ax+b^2x^2} - \sqrt{ax}}{b\sqrt{a}x^{3/2}} = \lim_{x \to 0^+} \frac{b^2x}{bx\sqrt{a}(\sqrt{1+b^2x/a}+1)\sqrt{x}/\sqrt{x}} = \frac{b}{2a} = 3 \implies b = 6a$$

Substitute $$b = 6a$$ into $$a+b=2 \implies 7a = 2 \implies a = \frac{2}{7}$$ and $$b = \frac{12}{7}$$.

$$\frac{b}{a} = \frac{12/7}{2/7} = 6$$