The value of `lim_(xrarroo)(x^(3)sin(1//x)-2x^(2))/(1+3x^(2))` is __________

Answer

To solve the limit \( \lim_{x \to \infty} \frac{x^3 \sin\left(\frac{1}{x}\right) - 2x^2}{1 + 3x^2} \), we will follow these steps:

### Step 1: Analyze the limit
As \( x \to \infty \), \( \frac{1}{x} \to 0 \). Therefore, we can use the small angle approximation for sine, which states that \( \sin(y) \approx y \) when \( y \) is close to 0. Thus, we have:
\[ \sin\left(\frac{1}{x}\right) \approx \frac{1}{x} \]

### Step 2: Substitute the approximation
Substituting this into the limit, we get:
\[ \lim_{x \to \infty} \frac{x^3 \left(\frac{1}{x}\right) - 2x^2}{1 + 3x^2} \]
This simplifies to:
\[ \lim_{x \to \infty} \frac{x^2 - 2x^2}{1 + 3x^2} = \lim_{x \to \infty} \frac{-x^2}{1 + 3x^2} \]

### Step 3: Simplify the expression
Now, we can simplify the fraction:
\[ \lim_{x \to \infty} \frac{-x^2}{1 + 3x^2} = \lim_{x \to \infty} \frac{-1}{\frac{1}{x^2} + 3} \]
As \( x \to \infty \), \( \frac{1}{x^2} \to 0 \). Therefore, the limit becomes:
\[ \lim_{x \to \infty} \frac{-1}{0 + 3} = \frac{-1}{3} \]

### Step 4: Conclusion
Thus, the value of the limit is:
\[ \boxed{-\frac{1}{3}} \]