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 of rational functions:

• If degree of numerator $\lt $ degree of denominator, limit is 0.

• If degree of numerator $\gt $ degree of denominator, limit is \(\pm\infty\).

• If degree of numerator = degree of denominator, limit is the ratio of leading coefficients.
This shortcut can save a lot of time.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to evaluate the limit of a rational function as \(x\) approaches infinity.

Step 2: Key Formula or Approach:
For a rational function \( \lim_{x\to\infty} \frac{P(x)}{Q(x)} \), where \(P(x)\) and \(Q(x)\) are polynomials, the limit is determined by the ratio of the leading terms (the terms with the highest power of x).
A more formal method is to divide both the numerator and the denominator by the highest power of \(x\) in the denominator.

Step 3: Detailed Explanation:
First, let's simplify the denominator of the expression:
Denominator = \(x^2 - 4x(1-x^2) = x^2 - 4x + 4x^3\).
So the limit is:
\[ \lim_{x\to\infty} \frac{4x^3 - x + 1}{4x^3 + x^2 - 4x} \]
The degree of the numerator polynomial is 3, and the degree of the denominator polynomial is also 3. Since the degrees are equal, the limit is the ratio of the leading coefficients.
Leading coefficient of numerator = 4.
Leading coefficient of denominator = 4.
\[ \text{Limit} = \frac{4}{4} = 1 \]

Formal Method:
The highest power of \(x\) in the expression is \(x^3\). Divide the numerator and denominator by \(x^3\):
\[ \lim_{x\to\infty} \frac{\frac{4x^3}{x^3} - \frac{x}{x^3} + \frac{1}{x^3}}{\frac{4x^3}{x^3} + \frac{x^2}{x^3} - \frac{4x}{x^3}} = \lim_{x\to\infty} \frac{4 - \frac{1}{x^2} + \frac{1}{x^3}}{4 + \frac{1}{x} - \frac{4}{x^2}} \]
As \(x \to \infty\), all terms with \(x\) in the denominator approach 0.
\[ \text{Limit} = \frac{4 - 0 + 0}{4 + 0 - 0} = \frac{4}{4} = 1 \]

Step 4: Final Answer:
The value of the limit is 1.