Let `f(x)=lim_(nto oo) (2x^(2n) sin (1/x)+x)/(1+x^(2n))` , then which of the following alternative(s) is/ are correct?

Options

A. `lim_(xto oo) xf(x)=2`
B. `lim_(xto1)f(x)` does not exist
C. `lim_(xto0) f(x)` does not exist
D. `lim_(xtooo) f(x)` is euqal to zero

Answer

To solve the problem, we need to evaluate the function \( f(x) = \lim_{n \to \infty} \frac{2x^{2n} \sin(1/x) + x}{1 + x^{2n}} \).

### Step-by-Step Solution:

1. **Identify the limit**:
We start with the expression:
\[ f(x) = \lim_{n \to \infty} \frac{2x^{2n} \sin(1/x) + x}{1 + x^{2n}} \]

2. **Factor out \( x^{2n} \)**:
We can factor \( x^{2n} \) from the numerator and denominator:
\[ f(x) = \lim_{n \to \infty} \frac{x^{2n} \left(2 \sin(1/x) + \frac{x}{x^{2n}}\right)}{x^{2n} \left(\frac{1}{x^{2n}} + 1\right)} \]
This simplifies to:
\[ f(x) = \lim_{n \to \infty} \frac{2 \sin(1/x) + \frac{x}{x^{2n}}}{\frac{1}{x^{2n}} + 1} \]

3. **Evaluate the limit as \( n \to \infty \)**:
As \( n \to \infty \), the term \( \frac{x}{x^{2n}} \) approaches 0 for \( x \neq 0 \), and \( \frac{1}{x^{2n}} \) also approaches 0 for \( x \neq 0 \). Thus, we have:
\[ f(x) = \frac{2 \sin(1/x)}{1} \]
Therefore:
\[ f(x) = 2 \sin(1/x) \]

4. **Check the behavior of \( f(x) \)**:
- As \( x \to 0 \), \( \sin(1/x) \) oscillates between -1 and 1, thus \( f(x) \) oscillates between -2 and 2.
- As \( x \to \infty \), \( \sin(1/x) \) approaches \( \sin(0) = 0 \), hence \( f(x) \to 0 \).

5. **Evaluate the options**:
- **Option 1**: \( \lim_{x \to \infty} x f(x) \) = \( \lim_{x \to \infty} x \cdot 2 \sin(1/x) \). As \( x \to \infty \), \( \sin(1/x) \to 0 \) and thus \( x f(x) \to 2 \).
- **Option 2**: \( \lim_{x \to 1} f(x) \) exists and equals \( 2 \sin(1) \).
- **Option 3**: \( f(x) \) does not exist at \( x = 0 \) because it oscillates.
- **Option 4**: \( \lim_{x \to \infty} f(x) = 0 \) exists.

### Conclusion:
- The correct options are:
- Option 1 is correct.
- Option 2 is correct.
- Option 3 is correct.
- Option 4 is correct.