Find the limit and show it does not exist () \( \lim _{x \rightarrow \infty} \frac{4 x+3}{3 x-1} \) (B) \( \lim _{x \rightarrow-\infty}\left(\sqrt{4 x^{2}+3 x}+2 x\right) \)

To find the limit for part (A): As \( x \) approaches infinity, we can simplify the expression \( \frac{4x + 3}{3x - 1} \) by dividing each term by \( x \): \[ \lim _{x \rightarrow \infty} \frac{4 + \frac{3}{x}}{3 - \frac{1}{x}} \] As \( x \) tends to infinity, the terms \( \frac{3}{x} \) and \( \frac{1}{x} \) both approach \( 0 \). Thus, the limit simplifies to: \[ \frac{4 + 0}{3 - 0} = \frac{4}{3} \] So the limit exists and is \( \frac{4}{3} \). For part (B): To evaluate \( \lim _{x \rightarrow -\infty}\left(\sqrt{4x^{2}+3x}+2x\right) \), we can factor out \( x^2 \) from under the square root: \[ \lim _{x \rightarrow -\infty} \left(\sqrt{x^2(4 + \frac{3}{x})} + 2x\right) = \lim _{x \rightarrow -\infty} \left(|x|\sqrt{4 + \frac{3}{x}} + 2x\right) \] Since \( x \) is approaching negative infinity, \( |x| = -x \): \[ = \lim _{x \rightarrow -\infty} \left(-x\sqrt{4 + \frac{3}{x}} + 2x\right) = \lim _{x \rightarrow -\infty} x\left(-\sqrt{4 + \frac{3}{x}} + 2\right) \] As \( x \rightarrow -\infty \), \( \frac{3}{x} \rightarrow 0 \), so: \[ = \lim _{x \rightarrow -\infty} x(-\sqrt{4} + 2) = \lim _{x \rightarrow -\infty} x(2 - 2) = \lim _{x \rightarrow -\infty} x(0) = 0 \] Thus, for part (B), the limit exists and equals \( 0 \). In summary, the limits are: (A) \( \frac{4}{3} \) exists, (B) \( 0 \) exists.