If f(x)=`{{:(,(x log cos x)/(log(1+x^(2))),x ne 0),(,0,x=0):}` then

Options

A. f(X) is not continuous at x=0
B. f(x) is continuous at x=0
C. f(x) is continuous at x=0 but not differentiable at x=0
D. f(x) is discontinuous at x=0

Answer

To determine the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \frac{x \log(\cos x)}{\log(1 + x^2)} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

we need to check if \( \lim_{x \to 0} f(x) = f(0) \).

### Step 1: Evaluate \( f(0) \)

From the definition of the function, we have:

\[ f(0) = 0 \]

### Step 2: Evaluate \( \lim_{x \to 0} f(x) \)

We need to find the limit of \( f(x) \) as \( x \) approaches 0. For \( x \neq 0 \):

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x \log(\cos x)}{\log(1 + x^2)} \]

### Step 3: Check the form of the limit

Substituting \( x = 0 \) directly into the limit gives us:

\[ \frac{0 \cdot \log(\cos(0))}{\log(1 + 0^2)} = \frac{0 \cdot \log(1)}{\log(1)} = \frac{0}{0} \]

This is an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule

Differentiate the numerator and denominator:

- **Numerator**: \( \frac{d}{dx}(x \log(\cos x)) \) using the product rule:

\[ \frac{d}{dx}(x \log(\cos x)) = \log(\cos x) + x \cdot \frac{d}{dx}(\log(\cos x)) = \log(\cos x) - x \tan x \]

- **Denominator**: \( \frac{d}{dx}(\log(1 + x^2)) \):

\[ \frac{d}{dx}(\log(1 + x^2)) = \frac{2x}{1 + x^2} \]

### Step 5: Rewrite the limit using derivatives

Now we rewrite the limit:

\[ \lim_{x \to 0} \frac{\log(\cos x) - x \tan x}{\frac{2x}{1 + x^2}} \]

This simplifies to:

\[ \lim_{x \to 0} \frac{(\log(\cos x) - x \tan x)(1 + x^2)}{2x} \]

### Step 6: Substitute \( x = 0 \) again

Substituting \( x = 0 \):

\[ \log(\cos(0)) = \log(1) = 0 \]
\[ \tan(0) = 0 \]

Thus, we have:

\[ \frac{(0 - 0)(1 + 0^2)}{2 \cdot 0} = \frac{0}{0} \]

### Step 7: Apply L'Hôpital's Rule again

Since we still have \( \frac{0}{0} \), we apply L'Hôpital's Rule again.

- Differentiate the numerator again:

\[ \frac{d}{dx}(\log(\cos x) - x \tan x) = -\tan x + \tan x + x \sec^2 x = x \sec^2 x \]

- Differentiate the denominator:

\[ \frac{d}{dx}(2x) = 2 \]

### Step 8: Rewrite the limit again

Now we have:

\[ \lim_{x \to 0} \frac{x \sec^2 x}{2} \]

### Step 9: Substitute \( x = 0 \)

Substituting \( x = 0 \):

\[ \frac{0 \cdot \sec^2(0)}{2} = \frac{0 \cdot 1}{2} = 0 \]

### Conclusion

Since:

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

Thus, \( f(x) \) is continuous at \( x = 0 \).

### Final Answer

The function \( f(x) \) is continuous at \( x = 0 \).

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