Let `f(x) = {{:({1 + |sin x|}^(a//|sin x|)",",-pi//6 lt x lt 0),(b",",x = 0),(e^(tan 2x//tan 3x)",",0 lt x lt pi//6):}` Determine a and b such that f(x) is continuous at x = 0

Answer

To determine the values of \( a \) and \( b \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit and the right-hand limit at \( x = 0 \) are equal to \( f(0) \).

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} \frac{(1 + |\sin x|)^a}{|\sin x|}, & -\frac{\pi}{6} < x < 0 \\ b, & x = 0 \\ \frac{e^{\tan(2x)}}{\tan(3x)}, & 0 < x < \frac{\pi}{6} \end{cases} \]

### Step 2: Calculate the left-hand limit as \( x \to 0^- \)
We need to find:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{(1 + |\sin x|)^a}{|\sin x|} \]
As \( x \to 0 \), \( |\sin x| \to 0 \). Therefore, we can rewrite the limit:
\[ \lim_{x \to 0^-} \frac{(1 + |\sin x|)^a}{|\sin x|} = \lim_{x \to 0^-} \frac{1^a}{|\sin x|} = \lim_{x \to 0^-} \frac{1}{|\sin x|} \text{ (since } |\sin x| \text{ approaches } 0) \]
This limit is of the form \( \frac{1}{0} \), which approaches \( \infty \).

However, we can use the expansion for small \( x \):
\[ |\sin x| \approx |x| \text{ as } x \to 0 \]
Thus, we can rewrite:
\[ \lim_{x \to 0^-} \frac{(1 + |x|)^a}{|x|} = \lim_{x \to 0^-} \frac{(1 + |x|)^a - 1}{|x|} + \lim_{x \to 0^-} \frac{1}{|x|} \]
Using the binomial approximation for small \( x \):
\[ (1 + |x|)^a \approx 1 + a|x| \text{ as } x \to 0 \]
Thus:
\[ \lim_{x \to 0^-} \frac{(1 + |x|)^a - 1}{|x|} = a \]
So:
\[ \lim_{x \to 0^-} f(x) = a \]

### Step 3: Calculate the right-hand limit as \( x \to 0^+ \)
Now we calculate:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{\tan(2x)}}{\tan(3x)} \]
Using the small angle approximation \( \tan(kx) \approx kx \):
\[ \lim_{x \to 0^+} \frac{e^{\tan(2x)}}{\tan(3x)} = \lim_{x \to 0^+} \frac{e^{2x}}{3x} = \frac{1}{3} \lim_{x \to 0^+} \frac{e^{2x}}{x} \]
As \( x \to 0 \), \( e^{2x} \to 1 \):
\[ = \frac{1}{3} \cdot \infty = \infty \]

### Step 4: Set the limits equal for continuity
For \( f(x) \) to be continuous at \( x = 0 \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]
This gives us:
\[ a = b \]
Since both limits approach infinity, we can conclude:
\[ b = a \]

### Step 5: Find specific values
To find specific values for \( a \) and \( b \), we can set:
\[ e^a = e^{\frac{2}{3}} \implies a = \frac{2}{3}, \quad b = e^{\frac{2}{3}} \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{2}{3}, \quad b = e^{\frac{2}{3}} \]