Question:

The value of 

\[ \lim_{x \to \infty} \frac{x^3 \sin\left(\frac{1}{x}\right) - 2x^2}{1 + 3x^2} \]

is ______.

Show Hint

For limits involving \( \sin(1/x) \), use small-angle approximation.
Updated On: Apr 30, 2026
  • \( 0 \)
  • \( \frac{1}{3} \)
  • \( -1 \)
  • \( -\frac{2}{3} \)
  • \( -\frac{1}{3} \)
Show Solution
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The Correct Option is

Solution and Explanation

Concept: Use: \[ \sin x \approx x \text{for small } x \]

Step 1:
Approximate sine. \[ \sin\left(\frac{1}{x}\right) \approx \frac{1}{x} \] \[ x^3 \sin\left(\frac{1}{x}\right) \approx x^2 \]

Step 2:
Simplify expression. \[ \frac{x^2 - 2x^2}{1 + 3x^2} = \frac{-x^2}{1+3x^2} \]

Step 3:
Divide by \(x^2\). \[ = \frac{-1}{\frac{1}{x^2} + 3} \] \[ \to \frac{-1}{3} \]
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