For the function `f (x) = lim _( x to 0 ) ( log (2 + x )-x ^(2) sin x )/( 1 + x ^(2 n )),` where of the following is true ? (n and x very independently),

Options

A. `lim _( x to 1 ^(-)) f (x)` does not exist
B. `lim _( x to 1 ^(+)) f (x)` does not exist
C. Both limits exist and `lim _( x to 10 ^(-)) f (x) = lim _( x to 1 ^(+)) f (x)`
D. Both limits exist and `lim _( x to 1 ^(-)) f (x) ne lim _( x to 1 ^(+)) f (x)`

Explanation

To solve the given limit function \( f(x) = \lim_{x \to 0} \frac{\log(2+x) - x^2 \sin x}{1 + x^{2n}} \), we will follow these steps:

### Step 1: Evaluate the limit as \( x \) approaches 0
We start by substituting \( x = 0 \) into the function.

\[ f(0) = \frac{\log(2+0) - 0^2 \sin(0)}{1 + 0^{2n}} = \frac{\log(2) - 0}{1 + 0} = \log(2) \]

### Step 2: Check if the limit is indeterminate
Next, we check if the limit is indeterminate. We need to evaluate the expression as \( x \) approaches 0.

- The numerator becomes:
\[ \log(2+x) - x^2 \sin x \to \log(2) - 0 = \log(2) \]

- The denominator becomes:
\[ 1 + x^{2n} \to 1 + 0 = 1 \]

Since both the numerator and denominator approach non-zero values, the limit is not indeterminate.

### Step 3: Conclusion of the limit
Thus, we conclude that:
\[ f(x) = \lim_{x \to 0} \frac{\log(2+x) - x^2 \sin x}{1 + x^{2n}} = \frac{\log(2)}{1} = \log(2) \]

### Step 4: Check the options
Now, we need to check the behavior of \( f(x) \) as \( x \) approaches 1 from both sides.

- As \( x \to 1^- \):
\[ f(1^-) = \log(2) \]

- As \( x \to 1^+ \):
\[ f(1^+) = \log(2) \]

Since both limits exist and are equal, we can conclude that the function is continuous at \( x = 1 \).

### Final Result
The function \( f(x) \) approaches \( \log(2) \) as \( x \) approaches 0, and it is continuous at \( x = 1 \).