2. \( \lim _{x \rightarrow-\infty} \frac{x-\sqrt{x^{2}+3 x}}{4 x+3} \)

To evaluate the limit \( \lim _{x \rightarrow-\infty} \frac{x-\sqrt{x^{2}+3 x}}{4 x+3} \), we start by simplifying the expression in the numerator: 1. Notice that as \( x \) approaches \( -\infty \), both \( x \) and \( \sqrt{x^2 + 3x} \) will be negative. To simplify the square root, factor out \( x^2 \) from inside: \[ \sqrt{x^2 + 3x} = \sqrt{x^2(1 + \frac{3}{x})} = |x|\sqrt{1 + \frac{3}{x}}. \] Since \( x \) is negative as it approaches \( -\infty \), \( |x| = -x \), so: \[ \sqrt{x^2 + 3x} = -x\sqrt{1 + \frac{3}{x}}. \] 2. Substitute this back into the limit: \[ \lim _{x \rightarrow-\infty} \frac{x - (-x\sqrt{1 + \frac{3}{x}})}{4x + 3} = \lim _{x \rightarrow-\infty} \frac{x + x\sqrt{1 + \frac{3}{x}}}{4x + 3}. \] Factoring \( x \) out of the numerator: \[ = \lim _{x \rightarrow-\infty} \frac{x(1 + \sqrt{1 + \frac{3}{x}})}{4x + 3}. \] 3. We can now factor \( x \) from the denominator as well: \[ = \lim _{x \rightarrow-\infty} \frac{1 + \sqrt{1 + \frac{3}{x}}}{4 + \frac{3}{x}}. \] 4. Now, as \( x \) approaches \( -\infty \), \( \frac{3}{x} \) approaches \( 0 \): \[ = \frac{1 + \sqrt{1 + 0}}{4 + 0} = \frac{1 + 1}{4} = \frac{2}{4} = \frac{1}{2}. \] Thus, the limit is: \[ \boxed{\frac{1}{2}}. \]