Giải thích các bước giải:

\(\begin{array}{l}
a.\mathop {\lim }\limits_{x \to 0} \frac{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to 0} \left( {{x^2} + 1} \right)\\
 = 0 + 1 = 1\\
b.\mathop {\lim }\limits_{x \to  - 1} \frac{{\sqrt {3{x^2} + 1}  - x}}{{x - 1}} = \frac{{\sqrt {3 + 1}  + 1}}{{ - 1 - 1}} =  - \frac{3}{2}\\
c.\mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + x + 1}}{{x - 2}}\\
 = \frac{{1 + 1 + 1}}{{1 - 2}} =  - 3\\
d.\mathop {\lim }\limits_{x \to 2} \frac{{4x + 1 - 9}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{4\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{4}{{x + 2}} = \frac{4}{4} = 1\\
e.\mathop {\lim }\limits_{x \to  + \infty } x\left( {\sqrt {1 + \frac{1}{x}}  + 1} \right) =  + \infty \\
Do:\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty \\
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {1 + \frac{1}{x}}  + 1} \right) = 2\\
f.\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} + x - {x^2}}}{{\sqrt {{x^2} + x}  + x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{x}{{\sqrt {{x^2} + x}  + x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{\sqrt {1 + \frac{1}{x}}  + 1}} = \frac{1}{2}
\end{array}\)

`lim_{x->0} (1+x+x^2+x^3)/(1+x)`

`= (1+0+0^2+0^3)/(1+0)`

`= 1`

`lim_{x->1} (x^3-1)/(x^2-3x+2)`

`= lim_{x->1} (x^2+x+1)/(x-2)`

`= (1^2+1+1)/(1-2)`

`= -3`

`lim_{x->+oo} (\sqrt{x^2+x} +x)`

`= lim_{x->+oo} (\sqrt{x^2+x}) + lim_{x->+oo}(x)`

`= +oo +oo`

`= +oo`

`lim_{x->-1} (\sqrt{3x^2+1}-x)/(x-1)`

`= (\sqrt{3(-1)^2 + 1} - (-1))/((-1)-1)`

`= -3/2`

`lim_{x->2} (\sqrt{4x+1}-3)/(x^2-4)`

`= lim_{x->2} ((4x-8)/(\sqrt{4x+1}+3) : x^2 - 4)`

`= lim_{x->2} ( 4/((\sqrt{4x+1}+3)(2+2)) )`

`= 4/((\sqrt{4.2+1}+3)(2+2))`

`= 1/6`

`lim_{x->+oo}(\sqrt{x^2+x}-x)`

`= lim_{x->+oo} x/(\sqrt{x^2+x}+x)`

`= lim_{x->+oo} 1/(\sqrt{1+1/x}+1)`

`= (lim_{x->+oo}(1))/(lim_{x->+oo}(\sqrt{1+1/x}+1))`

`= 1/2`