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

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

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

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

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

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

a: \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2x+8-16}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2}{\sqrt{2x+8}+4}=\dfrac{2}{\sqrt{2\cdot4+8}+4}\)

\(=\dfrac{2}{\sqrt{8+8}+4}=\dfrac{2}{4+4}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\dfrac{4x+1-9}{\sqrt{4x+1}+3}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{4\left(x-2\right)}\cdot\left(\sqrt{4x+1}+3\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}{4}\)

\(=\dfrac{\left(2+2\right)\left(\sqrt{4\cdot2+1}+3\right)}{4}=\sqrt{9}+3=6\)

c: \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\dfrac{4-x-2}{2+\sqrt{x+2}}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-x}\cdot\left(\sqrt{x+2}+2\right)\)

\(=\lim\limits_{x\rightarrow2}\left(-\sqrt{x+2}-2\right)\)

\(=-\sqrt{2+2}-2=-2-2=-4\)

a/ \(=\lim\limits_{x\rightarrow4^-}\dfrac{5-2x}{4-x}=\dfrac{-3}{0}=-\infty\)

b/ \(=\lim\limits_{x\rightarrow+\infty}x^3\left(-1+\dfrac{1}{x}-\dfrac{2}{x^2}+\dfrac{1}{x^3}\right)=-\infty\)

a/ \(=\lim\limits_{x\rightarrow4^-}\dfrac{5-2x}{4-x}=-\infty\)

b/ \(=\lim\limits_{x\rightarrow+\infty}x^3\left(-1\right)=-\infty\)

a: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x}-\sqrt[3]{x^2+7}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x}-2+2-\sqrt[3]{x^2+7}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{5-x-4}{\sqrt{5-x}+2}+\dfrac{8-x^2-7}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1-x}{\sqrt{5-x}+2}+\dfrac{1-x^2}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(1-x\right)\left(\dfrac{1}{\sqrt{5-x}+2}+\dfrac{1+x}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}\right)}{-\left(1-x\right)\left(1+x\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{\sqrt{5-x}+2}+\dfrac{1+x}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}}{-\left(1+x\right)}\)

\(=\dfrac{\dfrac{1}{\sqrt{5-1}+2}+\dfrac{1+1}{4+2\cdot\sqrt[3]{1^2+7}+\sqrt[3]{\left(1+7\right)^2}}}{-\left(1+1\right)}\)

\(=\dfrac{\dfrac{1}{2+1}+\dfrac{2}{4+2\cdot2+4}}{-2}\)

\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{6}}{-2}=-\dfrac{1}{4}\)

b: \(\lim\limits_{x\rightarrow4}\dfrac{x^2-4x}{x^2+x-20}\)

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

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

\(=\lim\limits_{x\rightarrow4}\dfrac{x}{x+5}=\dfrac{4}{4+5}=\dfrac{4}{9}\)

1: \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+1}-\sqrt{x+5}}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2x+1-x-5}{\sqrt{2x+1}+\sqrt{x+5}}\cdot\dfrac{1}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{x-4}{x-4}\cdot\dfrac{1}{\sqrt{2x+1}+\sqrt{x+5}}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{1}{\sqrt{2x+1}+\sqrt{x+5}}=\dfrac{1}{\sqrt{2\cdot4+1}+\sqrt{4+5}}\)

\(=\dfrac{1}{\sqrt{9}+\sqrt{9}}=\dfrac{1}{6}\)

2: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1-x}-\sqrt{1+x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{1-x-1-x}{\sqrt{1-x}+\sqrt{1+x}}\cdot\dfrac{1}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\cdot\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\sqrt{1-x}+\sqrt{1+x}}=\dfrac{-2}{\sqrt{1-0}+\sqrt{1+0}}\)
\(=\dfrac{-2}{1+1}=-1\)

a) Hàm số f(x) = xác định trên R\{} và ta có x = 4 ∈ (;+∞).

Giả sử (x_n) là dãy số bất kì và x_n ∈ (;+∞); x_n ≠ 4 và x_n → 4 khi n → +∞.

Ta có lim f(x_n) = lim = = .

Vậy = .

b) Hàm số f(x) = xác định trên R.

Giả sử (x_n) là dãy số bất kì và x_n → +∞ khi n → +∞.

Ta có lim f(x_n) = lim = lim = -5.

Vậy = -5.

\(\lim\limits_{x\rightarrow4}\frac{2x-\sqrt{3x+1}}{x^2-1}=\frac{8-\sqrt{11}}{15}\)

Nhưng mình đoán bạn ghi nhầm đề, x tiến tới 1 mới có lý

\(\lim\limits_{x\rightarrow1}\frac{2x-\sqrt{3x+1}}{x^2-1}=\lim\limits_{x\rightarrow1}\frac{4x^2-3x-1}{\left(x-1\right)\left(x+1\right)\left(2x+\sqrt{3x+1}\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(4x+1\right)}{\left(x-1\right)\left(x+1\right)\left(2x+\sqrt{3x+1}\right)}=\lim\limits_{x\rightarrow1}\frac{4x+1}{\left(x+1\right)\left(2x+\sqrt{3x+1}\right)}=\frac{5}{2\left(2+2\right)}=\frac{5}{8}\)

\(\lim\limits_{x\rightarrow8}\frac{\sqrt[3]{x}-2+2-\sqrt{x-4}}{x-8}=\lim\limits_{x\rightarrow8}\frac{\frac{x-8}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}-\frac{x-8}{2+\sqrt{x-4}}}{x-8}\)

\(=\lim\limits_{x\rightarrow8}\left(\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}-\frac{1}{2+\sqrt{x-4}}\right)=\frac{1}{12}-\frac{1}{4}=-\frac{1}{6}\)