Question

The value of \[ \lim_{x\to\infty}\frac{4x^3-x+1}{x^2-4x(1-x^2)} \] is

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(\infty\)

Hint:

For limits at infinity, compare the highest power terms in numerator and denominator.

Answer 1

The Correct Option is B

Step 1: Consider the denominator: \[ x^2-4x(1-x^2) \] \[ =x^2-4x+4x^3 \]

Step 2: Therefore, the limit becomes: \[ \lim_{x\to\infty}\frac{4x^3-x+1}{4x^3+x^2-4x} \]

Step 3: Divide numerator and denominator by \(x^3\): \[ \lim_{x\to\infty}\frac{4-\frac{1}{x^2}+\frac{1}{x^3}}{4+\frac{1}{x}-\frac{4}{x^2}} \]

Step 4: As \(x\to\infty\), terms containing \(\frac{1}{x}\) become zero: \[ \frac{4}{4}=1 \] \[ \boxed{1} \]