$\begin{array}{l}4)\quad \lim\limits_{x\to -\infty}\dfrac{\sqrt{2x^4 -x^2} - x\sqrt{x^2+3}}{x(5-2x)}\\ = \lim\limits_{x\to -\infty}\dfrac{x^2\sqrt{2 -\dfrac{1}{x^2}} - x|x|\sqrt{1+\dfrac{3}{x^2}}}{x^2\left(\dfrac{5}{x}-2\right)}\\ = \lim\limits_{x\to -\infty}\dfrac{x^2\sqrt{2 -\dfrac{1}{x^2}} +x^2\sqrt{1+\dfrac{3}{x^2}}}{x^2\left(\dfrac{5}{x}-2\right)}\\ = \lim\limits_{x\to -\infty}\dfrac{\sqrt{2 -\dfrac{1}{x^2}} +\sqrt{1+\dfrac{3}{x^2}}}{\dfrac{5}{x}-2}\\ = \dfrac{\sqrt{2 -0} + \sqrt{1 + 3.0}}{5.0-2}\\ =-\dfrac{1 + \sqrt2}{2}\\ 6)\quad \lim\limits_{x\to -\infty}\dfrac{3\sqrt{x^2 - 1} -\sqrt[3]{1-8x^3}}{6x+9}\\ = \lim\limits_{x\to -\infty}\dfrac{3|x|\sqrt{1 - \dfrac{1}{x^2}} -x\sqrt[3]{\dfrac{1}{x^3}-8}}{x\left(6+\dfrac{9}{x}\right)}\\ = \lim\limits_{x\to -\infty}\dfrac{-3x\sqrt{1 - \dfrac{1}{x^2}} -x\sqrt[3]{\dfrac{1}{x^3}-8}}{x\left(6+\dfrac{9}{x}\right)}\\ = \lim\limits_{x\to -\infty}\dfrac{-3\sqrt{1 - \dfrac{1}{x^2}} -\sqrt[3]{\dfrac{1}{x^3}-8}}{6+\dfrac{9}{x}}\\ = \dfrac{-3\sqrt{1-0} - \sqrt{0-8}}{6 + 9.0}\\ = -\dfrac16\\ 7)\quad \lim\limits_{x\to -\infty}\dfrac{(2x-1)\sqrt{x^2 -3}}{x-5x^2}\\ =\lim\limits_{x\to -\infty}\dfrac{x\left(2-\dfrac1x\right).|x|\sqrt{1 -\dfrac{3}{x^2}}}{x^2\left(\dfrac{1}{x}-5\right)}\\ =\lim\limits_{x\to -\infty}\dfrac{-x^2\left(2-\dfrac1x\right)\sqrt{1 -\dfrac{3}{x^2}}}{x^2\left(\dfrac{1}{x}-5\right)}\\ =\lim\limits_{x\to -\infty}\dfrac{\left(2-\dfrac1x\right)\sqrt{1 -\dfrac{3}{x^2}}}{\dfrac{1}{x}-5}\\ = \dfrac{(2 -0)\sqrt{1 - 3.0}}{0-5}\\ =-\dfrac25 \end{array}$
