Let `f(x)=lim_(n->oo)(log(2+x)-x^(2n)sinx)/(1+x^(2n))`. then

Options

A. f (x) is continuous at `x =1`
B. `lim _(xto1) f (x) =log, 3`
C. `lim _(xto 1 ^(+)) f (x) =-sin 1`
D. `lim _(x to 1 ^(+)) f (x)` does not exist

Answer

To solve the problem, we need to evaluate the limit:

\[ f(x) = \lim_{n \to \infty} \frac{\log(2+x) - x^{2n} \sin x}{1 + x^{2n}} \]

### Step 1: Analyze the limit as \( n \to \infty \)

We need to consider the behavior of \( x^{2n} \) as \( n \to \infty \) for different values of \( x \):

- If \( x < 1 \), then \( x^{2n} \to 0 \).
- If \( x = 1 \), then \( x^{2n} = 1^{2n} = 1 \).
- If \( x > 1 \), then \( x^{2n} \to \infty \).

### Step 2: Evaluate \( f(1) \)

For \( x = 1 \):

\[ f(1) = \lim_{n \to \infty} \frac{\log(2+1) - 1^{2n} \sin(1)}{1 + 1^{2n}} = \lim_{n \to \infty} \frac{\log(3) - \sin(1)}{2} \]

This simplifies to:

\[ f(1) = \frac{\log(3) - \sin(1)}{2} \]

### Step 3: Evaluate the left-hand limit as \( x \to 1^- \)

For \( x < 1 \):

\[ f(1-h) = \lim_{n \to \infty} \frac{\log(2 + (1-h)) - (1-h)^{2n} \sin(1-h)}{1 + (1-h)^{2n}} \]

As \( n \to \infty \), \( (1-h)^{2n} \to 0 \):

\[ f(1-h) = \lim_{n \to \infty} \frac{\log(3 - h) - 0}{1 + 0} = \log(3 - h) \]

Taking the limit as \( h \to 0 \):

\[ \lim_{h \to 0} f(1-h) = \log(3) \]

### Step 4: Evaluate the right-hand limit as \( x \to 1^+ \)

For \( x > 1 \):

\[ f(1+h) = \lim_{n \to \infty} \frac{\log(2 + (1+h)) - (1+h)^{2n} \sin(1+h)}{1 + (1+h)^{2n}} \]

As \( n \to \infty \), \( (1+h)^{2n} \to \infty \):

\[ f(1+h) = \lim_{n \to \infty} \frac{\log(3 + h) - \infty}{1 + \infty} = \lim_{n \to \infty} \frac{-\infty}{\infty} = -\sin(1+h) \cdot 0 = 0 \]

Taking the limit as \( h \to 0 \):

\[ \lim_{h \to 0} f(1+h) = -\sin(1) \]

### Step 5: Check continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \):

- Left-hand limit \( \lim_{x \to 1^-} f(x) = \log(3) \)
- Right-hand limit \( \lim_{x \to 1^+} f(x) = -\sin(1) \)
- Value at \( x = 1 \) is \( f(1) = \frac{\log(3) - \sin(1)}{2} \)

Since \( \log(3) \neq -\sin(1) \), the function is not continuous at \( x = 1 \).

### Conclusion

Thus, the function \( f(x) \) is not continuous at \( x = 1 \).