`-` 7)

`\lim_{x \to 2} (f(x) - 3x) = 0` 

`=>``\lim_{x \to 2} (f(x) - 3x) = 0` 

Do đó: `\lim_{x \to 2} f(x) = \lim_{x \to 2} 3x = 3(2) = 6`

 `-` Ta có:

`\lim_{x \to 2} \frac{[f(x) - 3x][f(x) + 2]}{x^2 - 4} = 3 \cdot \frac{\lim_{x \to 2} f(x) + 2}{\lim_{x \to 2} x + 2} = 3 \cdot \frac{6 + 2}{2 + 2} = 3 \cdot \frac{8}{4} = 3 \cdot 2 = 6`

`-> ` chọn C 

`-` 8)

`+)` ĐK: `sqrt{x^2+ax+b)-a=0` khi `x=0`

`=>sqrt{b}-a=0` `=>b=a^2`

`-` Thay `b=a^2` có:

`\lim_{x \to 0} \frac{(\sqrt{x^2 + ax + a^2} - a)(\sqrt{x^2 + ax + a^2} + a)(\sqrt{x + a} + \sqrt{a - x})}{(\sqrt{x + a} - \sqrt{a - x})(\sqrt{x + a} + \sqrt{a - x})(\sqrt{x^2 + ax + a^2} + a)} = 1`

`=>``\lim_{x \to 0} \frac{(x^2 + ax + a^2 - a^2)(\sqrt{x + a} + \sqrt{a - x})}{(x + a - (a - x))(\sqrt{x^2 + ax + a^2} + a)} = 1` 

`=>``\lim_{x \to 0} \frac{(x^2 + ax)(\sqrt{x + a} + \sqrt{a - x})}{2x(\sqrt{x^2 + ax + a^2} + a)} = 1`

`=>``\lim_{x \to 0} \frac{x(x + a)(\sqrt{x + a} + \sqrt{a - x})}{2x(\sqrt{x^2 + ax + a^2} + a)} = 1`

`=>` `\lim_{x \to 0} \frac{(x + a)(\sqrt{x + a} + \sqrt{a - x})}{2(\sqrt{x^2 + ax + a^2} + a)} = 1`

 `=>``\frac{(0 + a)(\sqrt{0 + a} + \sqrt{a - 0})}{2(\sqrt{0^2 + a(0) + a^2} + a)} = 1`

`=>`` \frac{a(\sqrt{a} + \sqrt{a})}{2(\sqrt{a^2} + a)} = 1`

`=>``\frac{a(2\sqrt{a})}{2(a + a)} = 1`

`=>`` \frac{2a\sqrt{a}}{4a} = 1`

`=>``\frac{\sqrt{a}}{2} = 1`

`=>``\sqrt{a} = 2`

`=>` `a=4` `->` `b = a^2 = 4^2 = 16`

`-` Vậy chọn C