Question

The value of \( \lim_{x\rightarrow\infty}\frac{x^{3}+2x^{2}\sin x-4x \cos x}{\sqrt{(3x^{2}+2x \cos x)^{3}}} = \)

• A. \( \frac{1}{\sqrt{3}} \)
• B. \( \frac{1}{9} \)
• C. \( \frac{1}{3\sqrt{3}} \)
• D. \( \frac{2}{3} \)

Hint:

For infinity-type limits, bounded oscillating terms like \( \sin x \) and \( \cos x \) can be treated as constants because they are completely dominated by high powers of \( x \) as \( x \) grows infinitely large.

Answer 1

The Correct Option is C

Concept: When evaluating limits as \( x \rightarrow \infty \), we identify the highest power of \( x \) in both the numerator and the denominator. We factor out this dominant power, reducing the remaining bounded oscillating components (like \( \sin x \) and \( \cos x \)) to zero via the squeeze theorem rule: \[ \lim_{x\rightarrow\infty} \frac{\sin x}{x} = 0, \quad \lim_{x\rightarrow\infty} \frac{\cos x}{x} = 0 \]

Step 1: Analyzing the highest degree power in the numerator.
The dominant term in the numerator is clearly \( x^3 \). Let us factor out \( x^3 \): \[ \text{Numerator} = x^3 \left( 1 + \frac{2\sin x}{x} - \frac{4\cos x}{x^2} \right) \]

Step 2: Analyzing the highest degree power in the denominator.
Inside the square root, the expression is cubed: \( (3x^2 + 2x\cos x)^3 \). Factoring out \( x^2 \) from inside the parentheses: \[ (3x^2 + 2x\cos x)^3 = \left[ x^2 \left( 3 + \frac{2\cos x}{x} \right) \right]^3 = x^6 \left( 3 + \frac{2\cos x}{x} \right)^3 \] Taking the square root: \[ \text{Denominator} = \sqrt{x^6 \left( 3 + \frac{2\cos x}{x} \right)^3} = x^3 \left( 3 + \frac{2\cos x}{x} \right)^{3/2} \]

Step 3: Evaluating the limit fraction.
Now divide the simplified numerator by the simplified denominator: \[ \lim_{x\rightarrow\infty} \frac{x^3 \left( 1 + \frac{2\sin x}{x} - \frac{4\cos x}{x^2} \right)}{x^3 \left( 3 + \frac{2\cos x}{x} \right)^{3/2}} = \frac{1 + 0 - 0}{(3 + 0)^{3/2}} = \frac{1}{3^{3/2}} = \frac{1}{3\sqrt{3}} \] This matches Option (C) perfectly.