- Đặt:

- Nên:

Ta có:

Ta có:

a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).

b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).

c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).

d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).

Lời giải:
1.

\(\lim\limits_{x\to -1}\frac{x^{2019}+1}{x^2+x}=\lim\limits_{x\to -1}\frac{(x+1)(x^{2018}-x^{2017}+x^{2016}-....-x+1)}{x(x+1)}=\lim\limits_{x\to -1}\frac{x^{2018}-x^{2017}+x^{2016}-....-x+1}{x}\)

\(=\frac{(-1)^{2018}-(-1)^{2017}+(-1)^{2016}+....-(-1)+1}{-1}\)

\(=\frac{\underbrace{1+1+....+1+1}_{2019}}{-1}=\frac{2019}{-1}=-2019\)

2.

\(\lim\limits_{x\to 1}\frac{(x-1)+(x^2-1)+(x^3-1)+....+(x^n-1)}{x-1}\\ =\lim\limits_{x\to 1}\frac{(x-1)+(x-1)(x+1)+(x-1)(x^2+x+1)+....+(x-1)(x^{n-1}+x^{n-2}+...+x+1)}{x-1}\)

$\lim\limits_{x\to 1}[1+(x+1)+(x^2+x+1)+....+(x^{n-1}+x^{n-2}+...+x+1)]$

$=1+2+3+....+n=n(n+1):2$

\(\)

\(=lim_{x->0}\left(\dfrac{1+x^2-1}{x^2\left(\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1\right)}\right)\)

\(=lim_{x->0}1=1\)

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{1 - 3{x^2}}}{{{x^2} + 2x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2}\left( {\frac{1}{{{x^2}}} - 3} \right)}}{{{x^2}\left( {1 + \frac{{2x}}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\frac{1}{{{x^2}}} - 3}}{{1 + \frac{2}{x}}} = \frac{{\mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{{x^2}}} - \mathop {\lim }\limits_{x \to  + \infty } 3}}{{\mathop {\lim }\limits_{x \to  + \infty } 1 + \mathop {\lim }\limits_{x \to  + \infty } \frac{2}{x}}} = \frac{{0 - 3}}{{1 + 0}} =  - 3\)

b) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{2}{{x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{2}{{x\left( {1 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{x}.\mathop {\lim }\limits_{x \to  - \infty } \frac{2}{{1 + \frac{1}{x}}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{x}.\frac{{\mathop {\lim }\limits_{x \to  - \infty } 2}}{{\mathop {\lim }\limits_{x \to  - \infty } 1 + \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{x}}} = 0.\frac{2}{{1 + 0}} = 0\).