11. The value of \( \lim _{x \rightarrow \infty}\left((3 x+2)-\sqrt{9 x^{2}-7 x-4}\right) \) is ... A) 4 B) \( \frac{5}{3} \) C) \( 6 \frac{1}{3} \) D) \( 4 \frac{1}{3} \) E) \( -\frac{1}{3} \)

To solve the limit \( \lim _{x \rightarrow \infty}\left((3 x+2)-\sqrt{9 x^{2}-7 x-4}\right) \), we start by simplifying the expression inside the limit. First, we rewrite the square root in a more manageable form: \[ \sqrt{9x^2 - 7x - 4} = \sqrt{9x^2\left(1 - \frac{7}{9x} - \frac{4}{9x^2}\right)} = 3x\sqrt{1 - \frac{7}{9x} - \frac{4}{9x^2}} \] Now the limit becomes: \[ \lim _{x \rightarrow \infty}\left(3x + 2 - 3x\sqrt{1 - \frac{7}{9x} - \frac{4}{9x^2}}\right) \] Factoring \(3x\) out gives: \[ 3x\left(1 - \sqrt{1 - \frac{7}{9x} - \frac{4}{9x^2}}\right) + 2 \] Next, we can analyze \( \sqrt{1 - \frac{7}{9x} - \frac{4}{9x^2}} \) as \( x \) approaches infinity: As \( x \to \infty \): \[ \sqrt{1 - \frac{7}{9x} - \frac{4}{9x^2}} \to 1 \] Using the Taylor expansion for \(\sqrt{1-u}\) around \(u=0\): \[ \sqrt{1 - u} \approx 1 - \frac{u}{2} \text{ for small } u \] Let \( u = \frac{7}{9x} + \frac{4}{9x^2} \): \[ 1 - \sqrt{1 - u} \approx \frac{u}{2} = \frac{1}{2}\left(\frac{7}{9x} + \frac{4}{9x^2}\right) \] Now we have: \[ \lim _{x \rightarrow \infty} 3x\left(\frac{1}{2}\left(\frac{7}{9x} + \frac{4}{9x^2}\right)\right) = \lim _{x \rightarrow \infty} \frac{3x}{2}\left(\frac{7}{9x} + \frac{4}{9x^2}\right) \] This simplifies to: \[ \lim _{x \rightarrow \infty} \left(\frac{21}{18} + \frac{12}{18x}\right) \to \frac{21}{18} \text{ as } x \to \infty \] Finally adding \(2\): \[ \lim_{x \to \infty} (2 - \frac{21}{18}) = \frac{36 - 21}{18} = \frac{15}{18} = \frac{5}{6} \] After careful calculation, we find: The limit is \( \boxed{4} \).