Let `f(x) = underset(n rarr oo)("Lim")( 2x^(2n) sin""(1)/(x) +x)/(1+x^(2n))` then find
`underset( x rarr -oo)("Lim") f(x)`

Answer

To solve the limit problem given by

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

we want to find

\[ \lim_{x \to -\infty} f(x). \]

### Step 1: Analyze the function as \( x \to -\infty \)

As \( x \) approaches \(-\infty\), we can analyze the behavior of each term in the function.

1. The term \( x^{2n} \) approaches \( +\infty \) since \( 2n \) is even.
2. The term \( \sin\left(\frac{1}{x}\right) \) approaches \( \sin(0) = 0 \) as \( x \to -\infty \).
3. The term \( 2x^{2n} \sin\left(\frac{1}{x}\right) \) will thus approach \( 0 \) because \( \sin\left(\frac{1}{x}\right) \) approaches \( 0 \).
4. The term \( x \) approaches \(-\infty\).

Thus, we can rewrite \( f(x) \) as:

\[ f(x) = \lim_{n \to \infty} \frac{0 + x}{1 + x^{2n}} = \lim_{n \to \infty} \frac{x}{1 + x^{2n}}. \]

### Step 2: Simplify the limit

Now, we need to analyze the limit of the expression as \( n \to \infty \):

\[ \lim_{n \to \infty} \frac{x}{1 + x^{2n}}. \]

As \( x \to -\infty \), \( x^{2n} \) approaches \( +\infty \) (since \( 2n \) is even), and thus \( 1 + x^{2n} \) also approaches \( +\infty \). Therefore, we can simplify the limit:

\[ \lim_{n \to \infty} \frac{x}{1 + x^{2n}} = \lim_{n \to \infty} \frac{x}{x^{2n}} \text{ (since \( x^{2n} \) dominates 1)}. \]

### Step 3: Further simplification

This can be rewritten as:

\[ \frac{x}{x^{2n}} = \frac{1}{x^{2n-1}}. \]

As \( n \to \infty \), \( x^{2n-1} \) approaches \( +\infty \) (because \( x \) is negative), leading to:

\[ \lim_{n \to \infty} \frac{1}{x^{2n-1}} = 0. \]

### Conclusion

Thus, we find that:

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

### Final Answer

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