Limit of \( \displaystyle \frac{4 \left(2 x - 1\right)^{2} - 1}{2 x + 2 \left(2 x - 1\right)^{2} - 1} \) as \( x \to \infty \)

Problem 1.65 · hard

1. \[ \lim_{x \to \infty}\left(\frac{4 \left(2 x - 1\right)^{2} - 1}{2 x + 2 \left(2 x - 1\right)^{2} - 1}\right) \]
   limitStarting with the original limit expression.✓ Proved
2. \[ = \lim_{x \to \infty}\left(\frac{16 x^{2} - 16 x + 3}{8 x^{2} - 6 x + 1}\right) \]
   algebra algebra simplifyExpand the squared term (2x - 1)^2. Distribute the constants into the parentheses. Combine like terms in the numerator and denominator.✓ Proved
3. \[ = \lim_{x \to \infty}\left(\frac{16 x^{2}}{8 x^{2} - 6 x + 1} - \frac{16 x}{8 x^{2} - 6 x + 1} + \frac{3}{8 x^{2} - 6 x + 1}\right) \]
   limit-lawSplit the limit into three separate parts using limit laws.✓ Proved
4. \[ = - \lim_{x \to \infty}\left(\frac{16 x}{8 x^{2} - 6 x + 1}\right) + \lim_{x \to \infty}\left(\frac{16 x^{2}}{8 x^{2} - 6 x + 1}\right) + \lim_{x \to \infty}\left(\frac{3}{8 x^{2} - 6 x + 1}\right) \]
   limit-lawApply the limit to each term individually.✓ Proved
5. \[ = \lim_{x \to \infty} 2 - \lim_{x \to \infty}\left(\frac{16 x}{8 x^{2} - 6 x + 1}\right) + \lim_{x \to \infty}\left(\frac{3}{8 x^{2} - 6 x + 1}\right) \]
   simplifyDivide the first term's numerator and denominator by x^2.✓ Proved
6. \[ = 2 - \lim_{x \to \infty}\left(\frac{16 x}{8 x^{2} - 6 x + 1}\right) + \lim_{x \to \infty}\left(\frac{3}{8 x^{2} - 6 x + 1}\right) \]
   simplifyEvaluate the limit of the first term.✓ Proved
7. \[ = 2 - \lim_{x \to \infty}\left(\frac{16}{8 x - 6 + \frac{1}{x}}\right) + \lim_{x \to \infty}\left(\frac{3}{8 x^{2} - 6 x + 1}\right) \]
   algebraDivide the second term's numerator and denominator by x.✓ Proved
8. \[ = 2 + \lim_{x \to \infty}\left(\frac{3}{8 x^{2} - 6 x + 1}\right) \]
   limitEvaluate the limit of the second term as x approaches infinity.✓ Proved
9. \[ = 2 \]
   limit simplifyEvaluate the limit of the third term as x approaches infinity. Final calculation.✓ Proved