\( \lim _{x \rightarrow -\infty } (\sqrt {4x^{2}+8x-5}+2x+b)=-5 \)

To evaluate the limit \[ \lim_{x \rightarrow -\infty} \left( \sqrt{4x^2 + 8x - 5} + 2x + b \right) = -5, \] we start by simplifying the expression inside the limit. First, we focus on the term \( \sqrt{4x^2 + 8x - 5} \). As \( x \) approaches \(-\infty\), we can factor out \( x^2 \) from the square root: \[ \sqrt{4x^2 + 8x - 5} = \sqrt{x^2(4 + \frac{8}{x} - \frac{5}{x^2})} = |x|\sqrt{4 + \frac{8}{x} - \frac{5}{x^2}}. \] Since \( x \) is approaching \(-\infty\), we have \( |x| = -x \). Thus, we can rewrite the expression as: \[ \sqrt{4x^2 + 8x - 5} = -x \sqrt{4 + \frac{8}{x} - \frac{5}{x^2}}. \] Next, we analyze the limit of the square root term as \( x \) approaches \(-\infty\): \[ \lim_{x \to -\infty} \sqrt{4 + \frac{8}{x} - \frac{5}{x^2}} = \sqrt{4 + 0 - 0} = \sqrt{4} = 2. \] Thus, we have: \[ \sqrt{4x^2 + 8x - 5} \sim -x \cdot 2 = -2x \quad \text{as } x \to -\infty. \] Now substituting this back into our limit expression, we get: \[ \sqrt{4x^2 + 8x - 5} + 2x + b \sim -2x + 2x + b = b. \] Therefore, the limit simplifies to: \[ \lim_{x \to -\infty} \left( \sqrt{4x^2 + 8x - 5} + 2x + b \right) = b. \] Setting this equal to \(-5\) gives us: \[ b = -5. \] Thus, the value of \( b \) is \[ \boxed{-5}. \]