\(\sqrt{a+12}-\sqrt[3]{81+63-19}=0\Rightarrow a=13\)

Khi đó

\(\dfrac{\sqrt{13x^2+4x+8}-\sqrt[3]{81x^2+63x-19}}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{\sqrt[]{13x^2+4x+8}-\left(3x+2\right)+\left(3x+2-\sqrt[3]{81x^2+83x-19}\right)}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{\dfrac{4\left(x-1\right)^2}{\sqrt[]{13x^2+4x+8}+\left(3x+2\right)}+\dfrac{27\left(x-1\right)^2\left(x+1\right)}{\left(3x+2\right)^2+\left(3x+2\right)\sqrt[3]{81x^2+63x-19}+\sqrt[3]{\left(81x^2+63x-19\right)^2}}}{\left(x-1\right)^2\left(x+1\right)}\)

a: \(\lim\limits_{x\rightarrow-2}\dfrac{4-x^2}{2x^2+7x+6}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(2-x\right)\left(2+x\right)}{2x^2+4x+3x+6}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(2-x\right)\left(x+2\right)}{\left(x+2\right)\left(2x+3\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{2-x}{2x+3}=\dfrac{2-\left(-2\right)}{2\cdot\left(-2\right)+3}=\dfrac{4}{-4+3}=-4\)

b: \(\lim\limits_{x\rightarrow4}\dfrac{2x^2-13x+20}{x^3+64}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2x^2-8x-5x+20}{\left(x+4\right)\left(x^2-4x+16\right)}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{\left(x-4\right)\left(2x-5\right)}{x^3+64}\)

\(=\dfrac{\left(4-4\right)\left(2\cdot4-5\right)}{4^3+64}=0\)

c: \(\lim\limits_{x\rightarrow-1}\dfrac{2x^2+8x+6}{-2x^2+7x+9}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{2x^2+2x+6x+6}{-2x^2-2x+9x+9}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x+6\right)}{-2x\left(x+1\right)+9\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x+6\right)}{\left(x+1\right)\left(-2x+9\right)}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{2x+6}{-2x+9}=\dfrac{2\cdot\left(-1\right)+6}{-2\cdot\left(-1\right)+9}\)

\(=\dfrac{4}{11}\)

\(4x^3-3x+1=\left(2x-1\right)^2\left(x+1\right)\) có nghiệm kép \(x=\dfrac{1}{2}\)

\(\Rightarrow\sqrt{1+ax^2}-bx-2=0\) có nhiều hơn 1 nghiệm \(x=\dfrac{1}{2}\)

\(\Rightarrow\sqrt{1+\dfrac{a}{4}}=\dfrac{b}{2}+2\Rightarrow\sqrt{a+4}=b+4\) (\(b\ge-4\))

\(\Rightarrow a=b^2+8b+12\)

\(\Rightarrow\sqrt{1+\left(b^2+8b+12\right)x^2}=bx+2\)

\(\Rightarrow1+\left(b^2+8b+12\right)x^2=b^2x^2+4bx+4\)

\(\Rightarrow\left(8b+12\right)x^2-4bx-3=0\)

\(\Rightarrow\left(2x-1\right)\left[\left(4b+6\right)x+3\right]=0\)

\(\Rightarrow\left(4b+6\right)x+3=0\) có nghiệm \(x=\dfrac{1}{2}\)

\(\Rightarrow2b+3+3=0\Rightarrow b=-3\) \(\Rightarrow a=-3\)

Khi đó:

\(\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12\left(2x-1\right)^2}{\left(x+1\right)\left(2x-1\right)^2\left(\sqrt{1-3x^2}+2-3x\right)}\)

\(=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12}{\left(x+1\right)\left(\sqrt{1-3x^2}+2-3x\right)}=-8\)

\(\Rightarrow c=-8\)

Lời giải:

\(\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{(x+1)(2x-1)^2}\)

Để giới hạn hàm đã cho hữu hạn thì $f(x)=\sqrt{1+ax^2}-bx-2$ có nhân tử là $(2x-1)^2$

$f(x)$ có nhân tử $2x-1 \Leftrightarrow f(\frac{1}{2})=0\Leftrightarrow b=\sqrt{4+a}-4$

Khi đó:

$\sqrt{1+ax^2}-bx-2=(2x-1)(2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}})$

Giờ ta cần xác định $a,b$ để $2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}}=0$ với $x=\frac{1}{2}$

$\Leftrightarrow \sqrt{4+a}=1\Leftrightarrow a=-3$

$b=\sqrt{4+a}-4=-3$

\(\lim\limits_{x\to 0,5}\frac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{-3(2x-1)^2(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(2x-1)^2(x+1)}\)

\(=\lim\limits_{x\to 0,5}\frac{-3(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(x+1)}=-2=c\)

Lời giải:
a) 

\(\lim\limits_{x\to +\infty}\frac{\sqrt[3]{x^3+2x^2-4x+1}}{\sqrt{2x^2+x-8}}=\lim\limits_{x\to +\infty}\frac{\sqrt[3]{1+\frac{2}{x}-\frac{4}{x^2}+\frac{1}{x^3}}}{\sqrt{2+\frac{1}{x}-\frac{8}{x^2}}}\)

\(=\frac{1}{\sqrt{2}}\)

b) 

\(\lim\limits_{x\to -\infty}\frac{\sqrt{x^2-2x+4}-x}{3x-1}=\lim\limits_{x\to -\infty}\frac{\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}+1}{-3+\frac{1}{x}}=\frac{-1}{3}\)

\(\lim\limits_{x\rightarrow-2}\dfrac{x^3+2x^2}{\sqrt{x^2+4x+4}}=\lim\limits_{x\rightarrow-2}\dfrac{x^2\left(x+2\right)}{\sqrt{\left(x+2\right)^2}}\)

\(=\lim\limits_{x\rightarrow-2}x^2=\left(-2\right)^2=4\)

p/s: bài này mình chưa học trên lớp nên ko chắc 100% đúng

\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x+1}}{\sqrt{x+\sqrt{x+1}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x^2}}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)

Câu c số 1 trong hay ngoài căn nhỉ?

`a)lim_{x->+oo}[5x^2+x^3+5]/[4x^3+1]`       `ĐK: 4x^3+1 ne 0`

`=lim_{x->+oo}[5/x+1+5/[x^3]]/[4+1/[x^3]]`

`=1/4`

`b)lim_{x->-oo}[2x^2-x+1]/[x^3+x-2x^2]`      `ĐK: x ne 0;x ne 1`

`=lim_{x->-oo}[2/x-1/[x^2]+1/[x^3]]/[1+1/[x^2]-2/x]`

`=0`

Câu `c` giống `b`.

a) (x⁴ – x² + x - 1) = x⁴(1 - ) = +∞.

b) (-2x³ + 3x² -5 ) = x³(-2 + ) = +∞.

c) = = +∞.

d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+x}{5-2x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left|x\right|\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)
 \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}+1}{\dfrac{5}{x}-2}=-1\).

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\sqrt{\dfrac{4x^2}{x^2}-\dfrac{2}{x^2}}-x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{1}{x^3}}}{-x\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4}-1}{-1-1}=\dfrac{3}{2}\)

b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2x}{x}+\dfrac{3}{x}}{-\sqrt{\dfrac{2x^2}{x^2}-\dfrac{3}{x^2}}}=\dfrac{2}{-\sqrt{2}}=-\sqrt{2}\)

c/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)