Question

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

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

Hint:

For limits as \(x\to\infty\), compare the highest powers of \(x\). If numerator and denominator have the same degree, the limit is the ratio of leading coefficients.

Answer 1

The Correct Option is B

We are given: \[ \lim_{x\to\infty}\frac{4x^3-x+1}{x^2-4x(1-x^2)}. \] First simplify the denominator: \[ x^2-4x(1-x^2). \] Expand: \[ x^2-4x+4x^3. \] Arrange in descending powers of \(x\): \[ 4x^3+x^2-4x. \] So the limit becomes: \[ \lim_{x\to\infty}\frac{4x^3-x+1}{4x^3+x^2-4x}. \] Now observe the highest power of \(x\) in numerator and denominator. The highest power is \[ x^3. \] Divide numerator and denominator by \(x^3\): \[ \lim_{x\to\infty}\frac{4-\frac{1}{x^2}+\frac{1}{x^3}}{4+\frac{1}{x}-\frac{4}{x^2}}. \] As \[ x\to\infty, \] we know \[ \frac{1}{x}\to 0,\quad \frac{1}{x^2}\to 0,\quad \frac{1}{x^3}\to 0. \] Therefore, \[ \lim_{x\to\infty}\frac{4-\frac{1}{x^2}+\frac{1}{x^3}}{4+\frac{1}{x}-\frac{4}{x^2}} = \frac{4}{4}. \] \[ =1. \] Hence, the value of the limit is \[ 1. \]