Key Concepts and Formulas

• Continuity of a function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
• Limit of $\tan x$: The standard limit $\lim_{x \to 0} \frac{\tan x}{x} = 1$.
• Algebraic manipulation of limits involving square roots: To evaluate limits of the form $\frac{\sqrt{A} - \sqrt{B}}{C}$, we often multiply the numerator and denominator by the conjugate, $\sqrt{A} + \sqrt{B}$.

Step-by-Step Solution

Step 1: Understand the condition for continuity. The problem states that the function $f(x)$ is continuous at $x=0$. This implies that the limit of the function as $x$ approaches 0 from the left must be equal to the function's value at $x=0$, and this must also be equal to the limit of the function as $x$ approaches 0 from the right. Mathematically, this is expressed as: $\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$ We are given $f(0) = 3$. Therefore, we must have: $\lim_{x \to 0^-} f(x) = 3 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 3$

Step 2: Evaluate the left-hand limit and form an equation. For $x < 0$, $f(x) = \frac{\tan ((a+1) x)+\mathrm{b} \tan x}{x}$. We need to find the limit as $x \to 0^-$ and set it equal to 3. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\tan ((a+1) x)+\mathrm{b} \tan x}{x}$ To evaluate this limit, we can use the standard limit $\lim_{x \to 0} \frac{\tan kx}{x} = k$. We can split the fraction into two terms: $\lim_{x \to 0^-} \left( \frac{\tan ((a+1) x)}{x} + \frac{\mathrm{b} \tan x}{x} \right)$ Applying the standard limit formula: $(a+1) \cdot \lim_{x \to 0^-} \frac{\tan ((a+1) x)}{(a+1)x} + b \cdot \lim_{x \to 0^-} \frac{\tan x}{x}$ Since $\lim_{x \to 0^-} \frac{\tan kx}{x} = k$ and $\lim_{x \to 0^-} \frac{\tan x}{x} = 1$: $(a+1) \cdot 1 + b \cdot 1 = a+1+b$ Since $f(x)$ is continuous at $x=0$, this limit must be equal to $f(0)=3$. $a+1+b = 3$ $a+b = 2 \quad \cdots (1)$

Step 3: Evaluate the right-hand limit and form another equation. For $x > 0$, $f(x) = \frac{\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2}-\sqrt{\mathrm{a} x}}{\mathrm{b} \sqrt{\mathrm{a}} x \sqrt{x}}$. We need to find the limit as $x \to 0^+$ and set it equal to 3. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2}-\sqrt{\mathrm{a} x}}{\mathrm{b} \sqrt{\mathrm{a}} x \sqrt{x}}$ Let's simplify the terms inside the square root in the numerator: $\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2} = \sqrt{x(\mathrm{a}+\mathrm{b}^2 x)} = \sqrt{x} \sqrt{\mathrm{a}+\mathrm{b}^2 x}$ Substituting this back into the limit expression: $\lim_{x \to 0^+} \frac{\sqrt{x} \sqrt{\mathrm{a}+\mathrm{b}^2 x}-\sqrt{x}\sqrt{\mathrm{a}}}{\mathrm{b} \sqrt{\mathrm{a}} x \sqrt{x}}$ We can factor out $\sqrt{x}$ from the numerator: $\lim_{x \to 0^+} \frac{\sqrt{x} (\sqrt{\mathrm{a}+\mathrm{b}^2 x}-\sqrt{\mathrm{a}})}{\mathrm{b} \sqrt{\mathrm{a}} x \sqrt{x}}$ Cancel $\sqrt{x}$ from the numerator and denominator: $\lim_{x \to 0^+} \frac{\sqrt{\mathrm{a}+\mathrm{b}^2 x}-\sqrt{\mathrm{a}}}{\mathrm{b} \sqrt{\mathrm{a}} x}$ This limit is of the indeterminate form $\frac{0}{0}$. To resolve this, we multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{\mathrm{a}+\mathrm{b}^2 x}+\sqrt{\mathrm{a}}$: $\lim_{x \to 0^+} \frac{(\sqrt{\mathrm{a}+\mathrm{b}^2 x}-\sqrt{\mathrm{a}})(\sqrt{\mathrm{a}+\mathrm{b}^2 x}+\sqrt{\mathrm{a}})}{\mathrm{b} \sqrt{\mathrm{a}} x (\sqrt{\mathrm{a}+\mathrm{b}^2 x}+\sqrt{\mathrm{a}})}$ Using the difference of squares formula $(A-B)(A+B) = A^2 - B^2$: $\lim_{x \to 0^+} \frac{(\mathrm{a}+\mathrm{b}^2 x) - \mathrm{a}}{\mathrm{b} \sqrt{\mathrm{a}} x (\sqrt{\mathrm{a}+\mathrm{b}^2 x}+\sqrt{\mathrm{a}})}$ Simplify the numerator: $\lim_{x \to 0^+} \frac{\mathrm{b}^2 x}{\mathrm{b} \sqrt{\mathrm{a}} x (\sqrt{\mathrm{a}+\mathrm{b}^2 x}+\sqrt{\mathrm{a}})}$ Cancel $x$ from the numerator and denominator (since $x \to 0^+$, $x \neq 0$): $\lim_{x \to 0^+} \frac{\mathrm{b}^2}{\mathrm{b} \sqrt{\mathrm{a}} (\sqrt{\mathrm{a}+\mathrm{b}^2 x}+\sqrt{\mathrm{a}})}$ Now, substitute $x=0$ into the expression: $\frac{\mathrm{b}^2}{\mathrm{b} \sqrt{\mathrm{a}} (\sqrt{\mathrm{a}+\mathrm{b}^2 (0)}+\sqrt{\mathrm{a}})} = \frac{\mathrm{b}^2}{\mathrm{b} \sqrt{\mathrm{a}} (\sqrt{\mathrm{a}}+\sqrt{\mathrm{a}})}$ $= \frac{\mathrm{b}^2}{\mathrm{b} \sqrt{\mathrm{a}} (2\sqrt{\mathrm{a}})} = \frac{\mathrm{b}^2}{2 \mathrm{b} \mathrm{a}}$ Simplify the expression: $\frac{\mathrm{b}}{2 \mathrm{a}}$ Since $f(x)$ is continuous at $x=0$, this limit must be equal to $f(0)=3$. $\frac{\mathrm{b}}{2 \mathrm{a}} = 3$ $\frac{\mathrm{b}}{\mathrm{a}} = 6 \quad \cdots (2)$

Step 4: Solve the system of equations for the required ratio. We have two equations: (1) $a+b = 2$ (2) $\frac{b}{a} = 6$

From equation (2), we can express $b$ in terms of $a$: $b = 6a$

Substitute this expression for $b$ into equation (1): $a + (6a) = 2$ $7a = 2$ $a = \frac{2}{7}$

Now, substitute the value of $a$ back into the expression for $b$: $b = 6a = 6 \left(\frac{2}{7}\right) = \frac{12}{7}$

We are asked to find the value of $\frac{b}{a}$. We already found this in equation (2). $\frac{b}{a} = 6$

Common Mistakes & Tips

• Algebraic errors with square roots: Be careful when rationalizing the numerator involving square roots. Ensure the conjugate is applied correctly and the difference of squares formula is used properly.
• Canceling terms prematurely: Ensure that any term being canceled (like $x$ or $\sqrt{x}$) is not zero at the limit point. For limits as $x \to 0^+$, $x$ is positive and non-zero.
• Applying standard limits: Remember the standard limit $\lim_{x \to 0} \frac{\tan kx}{x} = k$. Incorrect application of this can lead to errors.

Summary

The problem requires us to use the condition of continuity of the function at $x=0$, which means the left-hand limit, the right-hand limit, and the function value at $x=0$ must all be equal. We calculated the left-hand limit using the standard tangent limit, which gave us the equation $a+b=2$. We calculated the right-hand limit by rationalizing the numerator and simplifying, which resulted in the equation $\frac{b}{a}=6$. Solving these two equations simultaneously allowed us to determine the value of $\frac{b}{a}$.

The final answer is $\boxed{6}$.