What is the limit as X approaches Infinity, x^2 -4 / (2+x-4x^2)

We need to find the limit of the expression as \( x \to \infty \):

\[
\lim_{x \to \infty} \frac{x^2 - 4}{2 + x - 4x^2}
\]

### Step-by-Step Solution:

1. **Write the expression:**
   \[
   \frac{x^2 - 4}{2 + x - 4x^2}
   \]

2. **Identify the highest powers of \( x \):**
   The highest power of \( x \) in the numerator is \( x^2 \), and the highest power of \( x \) in the denominator is also \( x^2 \).

3. **Divide every term by \( x^2 \) to simplify the expression:**
   \[
   \frac{\frac{x^2}{x^2} - \frac{4}{x^2}}{\frac{2}{x^2} + \frac{x}{x^2} - \frac{4x^2}{x^2}} = \frac{1 - \frac{4}{x^2}}{\frac{2}{x^2} + \frac{1}{x} - 4}
   \]

4. **Simplify as \( x \to \infty \):**
   As \( x \to \infty \), the terms \( \frac{4}{x^2} \), \( \frac{2}{x^2} \), and \( \frac{1}{x} \) tend to 0.

   Thus, the expression becomes:
   \[
   \frac{1 - 0}{0 + 0 - 4} = \frac{1}{-4}
   \]

5. **Conclusion:**
   The limit is:
   \[
   \lim_{x \to \infty} \frac{x^2 - 4}{2 + x - 4x^2} = -\frac{1}{4}
   \]

Would you like more details or explanations on any steps?

Here are 5 related questions you might explore:

1. What is the limit of \( \frac{x^2}{x^2 - 1} \) as \( x \to \infty \)?
2. How do you evaluate limits involving polynomials and rational functions?
3. What happens to \( \frac{1}{x} \) as \( x \to \infty \)?
4. Can we always divide by the highest power of \( x \) when finding limits at infinity?
5. What is the limit of \( \frac{3x^3 + 2x^2 - x}{5x^3 - x + 4} \) as \( x \to \infty \)?

**Tip:** When dealing with limits at infinity, focus on the highest-degree terms in the numerator and denominator, as they dominate the behavior of the function.

What is the limit as X approaches Infinity, x^2 - 4 / (2 + x - 4x^2)?

Explore how to solve the limit of the rational function (x^2 - 4) / (2 + x - 4x^2) as x approaches infinity. This step-by-step solution simplifies the expression and uses the Limit Theorem at Infinity to arrive at the result of -1/4. Suitable for advanced high school and college-level students.

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