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:

\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]

Given the piecewise function:

\[ f(x) = \begin{cases} (1 + |\sin x|)^{\frac{a}{|\sin x|}}, & -\frac{\pi}{6} < x < 0 \\ b, & x = 0 \\ e^{\frac{\tan 2x}{\tan 3x}}, & 0 < x < \frac{\pi}{6} \end{cases} \]

### Step 1: Calculate \( \lim_{x \to 0^-} f(x) \)

For \( x \to 0^- \):

\[ f(x) = (1 + |\sin x|)^{\frac{a}{|\sin x|}} \]

As \( x \to 0 \), \( |\sin x| \to 0 \). Thus, we can rewrite the limit:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0} (1 + |\sin x|)^{\frac{a}{|\sin x|}} \]

This is an indeterminate form \( 1^{\infty} \). To resolve this, we can take the natural logarithm:

Let \( y = (1 + |\sin x|)^{\frac{a}{|\sin x|}} \)

Taking the logarithm:

\[ \ln y = \frac{a}{|\sin x|} \ln(1 + |\sin x|) \]

Now, as \( x \to 0 \), \( |\sin x| \to 0 \) and we can use the approximation \( \ln(1 + u) \approx u \) for small \( u \):

\[ \ln(1 + |\sin x|) \approx |\sin x| \quad \text{as } x \to 0 \]

Thus,

\[ \ln y \approx \frac{a}{|\sin x|} |\sin x| = a \]

So,

\[ \lim_{x \to 0^-} \ln y = a \implies \lim_{x \to 0^-} y = e^a \]

Thus,

\[ \lim_{x \to 0^-} f(x) = e^a \]

### Step 2: Calculate \( f(0) \)

From the definition of the function,

\[ f(0) = b \]

### Step 3: Calculate \( \lim_{x \to 0^+} f(x) \)

For \( x \to 0^+ \):

\[ f(x) = e^{\frac{\tan 2x}{\tan 3x}} \]

As \( x \to 0 \), both \( \tan 2x \) and \( \tan 3x \) approach 0. We can use L'Hôpital's rule:

\[ \lim_{x \to 0^+} \frac{\tan 2x}{\tan 3x} = \lim_{x \to 0^+} \frac{2\sec^2(2x)}{3\sec^2(3x)} = \frac{2}{3} \]

Thus,

\[ \lim_{x \to 0^+} f(x) = e^{\frac{2}{3}} \]

### Step 4: Set the limits equal for continuity

For continuity at \( x = 0 \):

\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]

This gives us:

\[ e^a = b = e^{\frac{2}{3}} \]

### Step 5: Solve for \( a \) and \( b \)

From \( e^a = e^{\frac{2}{3}} \), we have:

\[ a = \frac{2}{3} \]

And since \( b = e^{\frac{2}{3}} \), we have:

\[ b = e^{\frac{2}{3}} \]

### Final Answer

Thus, the values of \( a \) and \( b \) are:

\[ a = \frac{2}{3}, \quad b = e^{\frac{2}{3}} \]