Step 1: Understanding the Concept:
For a piecewise function to be continuous at a specific point (here, $x=0$), the left-hand limit (LHL) approaching the point, the right-hand limit (RHL) approaching the point, and the actual function value at that point must all exist and be exactly equal. We will calculate the limits using standard trigonometric and exponential limit formulas.
Step 2: Key Formula or Approach:
1. Condition for continuity at $x=c$: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
2. Standard Limits to use as $x \to 0$:
- $\lim_{x \to 0} \frac{\sin x}{x} = 1$
- $\lim_{x \to 0} \frac{\tan x}{x} = 1$
- $\lim_{x \to 0} \frac{k^x - 1}{x} = \log_e k$
Step 3: Detailed Explanation:
Let's find the Left-Hand Limit (LHL) as $x \to 0^-$. We use the top branch function:
\[ \text{LHL} = \lim_{x \to 0^-} \frac{3\sin x + 5\tan x}{a^x - 1} \]
Divide the numerator and denominator by $x$ to create standard limit forms:
\[ \text{LHL} = \lim_{x \to 0^-} \frac{\frac{3\sin x}{x} + \frac{5\tan x}{x}}{\frac{a^x - 1}{x}} \]
Applying the standard limits:
\[ \text{LHL} = \frac{3(1) + 5(1)}{\log_e a} = \frac{8}{\log_e a} \]
Now let's find the Right-Hand Limit (RHL) as $x \to 0^+$. We use the bottom branch function:
\[ \text{RHL} = \lim_{x \to 0^+} \frac{8x + 2x\cos x}{b^x - 1} \]
Factor out $x$ in the numerator and then divide numerator and denominator by $x$:
\[ \text{RHL} = \lim_{x \to 0^+} \frac{x(8 + 2\cos x)}{b^x - 1} = \lim_{x \to 0^+} \frac{8 + 2\cos x}{\frac{b^x - 1}{x}} \]
As $x \to 0$, $\cos x \to 1$. Apply the standard exponential limit:
\[ \text{RHL} = \frac{8 + 2(1)}{\log_e b} = \frac{10}{\log_e b} \]
The function value at $x=0$ is given explicitly in the middle branch:
\[ f(0) = \frac{2}{\log_e 2} \]
For continuity, LHL = RHL = $f(0)$. Let's set up the equations.
Equation 1 (LHL = $f(0)$):
\[ \frac{8}{\log_e a} = \frac{2}{\log_e 2} \]
Cross-multiply and solve for $\log_e a$:
\[ 2 \log_e a = 8 \log_e 2 \implies \log_e a = 4 \log_e 2 \]
Using the power property $\log(m^n) = n\log m$:
\[ \log_e a = \log_e (2^4) = \log_e 16 \implies a = 16 \]
Equation 2 (RHL = $f(0)$):
\[ \frac{10}{\log_e b} = \frac{2}{\log_e 2} \]
Cross-multiply and solve for $\log_e b$:
\[ 2 \log_e b = 10 \log_e 2 \implies \log_e b = 5 \log_e 2 \]
\[ \log_e b = \log_e (2^5) = \log_e 32 \implies b = 32 \]
Step 4: Final Answer:
The values are 16, 32.