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 limit function defined as:

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

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

In the expression, both the numerator and denominator contain terms that grow with \( n \). The highest power of \( x \) in both the numerator and denominator is \( x^{2n} \).

### Step 2: Factor out \( x^{2n} \)

We can factor \( x^{2n} \) from both the numerator and the denominator:

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

This simplifies to:

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

### Step 3: Evaluate the limit as \( n \to \infty \)

As \( n \to \infty \), \( \frac{x}{x^{2n}} \) approaches 0 (for \( x \neq 0 \)) and \( \frac{1}{x^{2n}} \) also approaches 0. Therefore, we have:

\[ f(x) = \frac{2 \sin\left(\frac{1}{x}\right)}{1} \]

Thus, we can write:

\[ f(x) = 2 \sin\left(\frac{1}{x}\right) \]

### Step 4: Check the options

1. **Option 1: \( \lim_{x \to \infty} x f(x) = 2 \)**

We compute:

\[ \lim_{x \to \infty} x f(x) = \lim_{x \to \infty} x \cdot 2 \sin\left(\frac{1}{x}\right) \]

Using the fact that \( \sin\left(\frac{1}{x}\right) \approx \frac{1}{x} \) as \( x \to \infty \):

\[ \lim_{x \to \infty} x \cdot 2 \cdot \frac{1}{x} = 2 \]

Thus, **Option 1 is correct**.

2. **Option 2: \( \lim_{x \to 1} f(x) \) does not exist**

We calculate:

\[ f(1) = 2 \sin(1) \]

Since \( \sin(1) \) is defined, **Option 2 is incorrect**.

3. **Option 3: \( \lim_{x \to 0} f(x) \) does not exist**

As \( x \to 0 \):

\[ f(x) = 2 \sin\left(\frac{1}{x}\right) \]

The sine function oscillates between -1 and 1, hence \( f(x) \) does not converge. Thus, **Option 3 is correct**.

4. **Option 4: \( \lim_{x \to \infty} f(x) = 0 \)**

As \( x \to \infty \):

\[ f(x) = 2 \sin\left(\frac{1}{x}\right) \to 2 \cdot 0 = 0 \]

Thus, **Option 4 is correct**.

### Conclusion

The correct options are **1, 3, and 4**.