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

To find the limit, we can start by simplifying the expression. As \( x \) approaches \(-\infty\), the leading term in the square root will dominate. The expression can be rewritten as: \[ \lim_{x \rightarrow -\infty} \frac{3x}{\sqrt{4x^2 - 2x + 2}} = \lim_{x \rightarrow -\infty} \frac{3x}{\sqrt{4x^2(1 - \frac{2}{4x} + \frac{2}{4x^2})}} = \lim_{x \rightarrow -\infty} \frac{3x}{\sqrt{4x^2} \sqrt{1 - \frac{2}{4x} + \frac{2}{4x^2}}} \] Now, calculate \( \sqrt{4x^2} \): \[ \sqrt{4x^2} = 2|x| = 2(-x) = -2x \quad \text{(since \( x \) is negative)} \] Substituting this back into the limit, we have: \[ \lim_{x \rightarrow -\infty} \frac{3x}{-2x \sqrt{1 - \frac{2}{4x} + \frac{2}{4x^2}}} \] This simplifies to: \[ \lim_{x \rightarrow -\infty} \frac{3}{-2 \sqrt{1 - \frac{2}{4x} + \frac{2}{4x^2}}} \] As \( x \) approaches \(-\infty\), \( -\frac{2}{4x} \) and \( \frac{2}{4x^2} \) trend towards 0. Thus, we have: \[ \sqrt{1 - 0 + 0} = 1 \] Finally, the limit becomes: \[ \frac{3}{-2} = -\frac{3}{2} \] Thus, the limit is: \[ \lim_{x \rightarrow -\infty} \frac{3x}{\sqrt{4x^2 - 2x + 2}} = -\frac{3}{2} \]