\( \begin{array}{lll}\text { (a) } \lim _{x \rightarrow-\infty} \frac{3 x^{6}-7 x^{5}+x}{5 x^{6}+4 x^{5}-3} & \text { (b) } \lim _{x \rightarrow \infty} \frac{5 x+2 x^{3}}{x^{3}+x-7} & \text { (c) } \lim _{x \rightarrow-\infty} \frac{\sqrt{4 x^{2}+3 x}-7}{7-3 x} \\ \text { (d) } \lim _{x \rightarrow \infty} \frac{x^{2}}{e^{4 x}-1-4 x} & \text { (e) } \lim _{x \rightarrow \infty} \frac{1+3 x}{\sqrt{2 x^{2}+x}} & \text { (f) } \lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-\sqrt{x^{2}+1}\right)\end{array} \)

(a) We have \[ \lim_{x\rightarrow-\infty}\frac{3x^6-7x^5+x}{5x^6+4x^5-3}. \] Factor out \(x^6\) in both numerator and denominator: \[ \frac{x^6\left(3-7/x+1/x^5\right)}{x^6\left(5+4/x-3/x^6\right)} = \frac{3-7/x+1/x^5}{5+4/x-3/x^6}. \] Taking the limit as \(x\rightarrow-\infty\) (where the terms with \(1/x\) vanish), we obtain \[ \frac{3}{5}. \] --- (b) We have \[ \lim_{x\rightarrow \infty}\frac{5x+2x^3}{x^3+x-7}. \] Divide numerator and denominator by \(x^3\): \[ \frac{\frac{5}{x^2}+2}{1+\frac{1}{x^2}-\frac{7}{x^3}}. \] As \(x\rightarrow\infty\), the terms with \(1/x^2\) and \(1/x^3\) vanish, so the limit becomes \[ \frac{2}{1}=2. \] --- (c) We have \[ \lim_{x\rightarrow-\infty}\frac{\sqrt{4x^2+3x}-7}{7-3x}. \] First, factor inside the square root: \[ \sqrt{4x^2+3x} = \sqrt{4x^2\left(1+\frac{3}{4x}\right)} = 2|x|\sqrt{1+\frac{3}{4x}}. \] For \(x\rightarrow -\infty\), we have \(|x|=-x\), so \[ \sqrt{4x^2+3x} = -2x\sqrt{1+\frac{3}{4x}}. \] Thus, the expression becomes \[ \frac{-2x\sqrt{1+\frac{3}{4x}}-7}{7-3x}. \] For very large negative \(x\), the dominant term in the numerator is \(-2x\sqrt{1+\frac{3}{4x}} \approx -2x\) and in the denominator it is \(-3x\). Therefore, \[ \lim_{x\rightarrow-\infty} \frac{-2x}{-3x}=\frac{2}{3}. \] --- (d) We have \[ \lim_{x\rightarrow\infty}\frac{x^2}{e^{4x}-1-4x}. \] Since \(e^{4x}\) grows exponentially while \(x^2\) grows polynomially, the denominator becomes infinitely large compared to the numerator. Hence, the limit is \[ 0. \] --- (e) We have \[ \lim_{x\rightarrow\infty}\frac{1+3x}{\sqrt{2x^2+x}}. \] Divide numerator and denominator by \(x\). In the numerator, factor out \(x\) (for large \(x\), \(1\) is negligible): \[ 1+3x \sim 3x. \] In the denominator, write \[ \sqrt{2x^2+x} = x\sqrt{2+\frac{1}{x}} \sim x\sqrt{2}. \] Thus, \[ \lim_{x\rightarrow\infty} \frac{3x}{x\sqrt{2}}=\frac{3}{\sqrt{2}}. \] --- (f) We have \[ \lim_{x\rightarrow\infty}\left(\sqrt{x^2-x+1}-\sqrt{x^2+1}\right). \] Rationalize the expression by multiplying and dividing by the conjugate: \[ \sqrt{x^2-x+1}-\sqrt{x^2+1} = \frac{(\sqrt{x^2-x+1}-\sqrt{x^2+1})(\sqrt{x^2-x+1}+\sqrt{x^2+1})}{\sqrt{x^2-x+1}+\sqrt{x^2+1}}. \] The numerator becomes \[ (x^2-x+1) - (x^2+1) = -x. \] Thus, the expression is \[ \frac{-x}{\sqrt{x^2-x+1}+\sqrt{x^2+1}}. \] For large \(x\), \[ \sqrt{x^2-x+1} \sim x \quad \text{and} \quad \sqrt{x^2+1} \sim x, \] so the denominator is approximately \(2x\). Therefore, \[ \lim_{x\rightarrow\infty}\frac{-x}{2x}=-\frac{1}{2}. \]