1.
     $\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt{1-2x}-\sqrt[3]{1+3x}}{x^2}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{(\sqrt{1-2x}-1)-(\sqrt[3]{1+3x}-1)}{x^2}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{-2x}{\sqrt{1-2x}+1}-\dfrac{3x}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}}{x^2}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{-2}{\sqrt{1-2x}+1}-\dfrac{3}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}}{x}$
Mà $\mathop {\lim }\limits_{x \to 0^+}\dfrac{\dfrac{-2}{\sqrt{1-2x}+1}-\dfrac{3}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}}{x}=-\infty$
       $\mathop {\lim }\limits_{x \to 0^-}\dfrac{\dfrac{-2}{\sqrt{1-2x}+1}-\dfrac{3}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}}{x}=+\infty$
nên giới hạn cần tìm không tồn tại.

b.
     $\mathop {\lim }\limits_{x \to 1}\dfrac{\sqrt{3x+6}-\sqrt[3]{24+3x}}{x^2-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{(\sqrt{3x+6}-3)-(\sqrt[3]{24+3x}-3)}{x^2-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{\dfrac{3x-3}{\sqrt{3x+6}+3}-\dfrac{3x-3}{\sqrt[3]{(24+3x)^2}+3\sqrt[3]{24+3x}+9}}{x^2-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{\dfrac{3}{\sqrt{3x+6}+3}-\dfrac{3}{\sqrt[3]{(24+3x)^2}+3\sqrt[3]{24+3x}+9}}{x+1}=\dfrac{7}{36}$

a.
     $\mathop {\lim }\limits_{x \to 2}\dfrac{x^3-\sqrt{3x+58}}{x-2}$
$=\mathop {\lim }\limits_{x \to 2}\dfrac{x^6-(3x+58)}{(x-2)(x^3+\sqrt{3x+58})}$
$=\mathop {\lim }\limits_{x \to 2}\dfrac{(x-2)(x^5+2x^4+4x^3+8x^2+16x+29)}{(x-2)(x^3+\sqrt{3x+58})}$
$=\mathop {\lim }\limits_{x \to 2}\dfrac{x^5+2x^4+4x^3+8x^2+16x+29}{x^3+\sqrt{3x+58}}=\dfrac{189}{16}$

8.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{2x^2+x-1}{x\sqrt{x^2-1}}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{2+\dfrac{1}{x}-\dfrac{1}{x^2}}{-\sqrt{1-\dfrac{1}{x^2}}}=-2$

7.
      $\mathop {\lim }\limits_{x \to +\infty}\dfrac{x+\sqrt{x^2+1}}{2x+\sqrt{x^2+1}}$
$=\mathop {\lim }\limits_{x \to +\infty}\dfrac{1+\sqrt{1+\dfrac{1}{x^2}}}{2+\sqrt{1+\dfrac{1}{x^2}}}=\dfrac{2}{3}$

6.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{\sqrt{2x^4+x^2-1}}{1-2x}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{\dfrac{\sqrt{2x^4+x^2-1}}{|x|}}{\dfrac{1-2x}{|x|}}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{\sqrt{2x^2+1-\dfrac{1}{x^2}}}{\dfrac{1}{|x|}+2}=+\infty$

5.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{(x-1)(x-2)(x-3)(x-4)(x-5)}{(5x-1)^5}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{(1-\dfrac{1}{x})(1-\dfrac{2}{x})(1-\dfrac{3}{x})(1-\dfrac{4}{x})(1-\dfrac{5}{x})}{(5-\dfrac{1}{x})^5}=\dfrac{1}{5^5}$

4.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{\sqrt{x^2+2x}+3x}{\sqrt{4x^2+1}-x+2}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{\dfrac{\sqrt{x^2+2x}+3x}{|x|}}{\dfrac{\sqrt{4x^2+1}-x+2}{|x|}}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{\sqrt{1+\dfrac{2}{x}}-3}{\sqrt{4+\dfrac{1}{x^2}}+1+\dfrac{2}{|x|}}=\dfrac{-2}{3}$