a) \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right) = \mathop {\lim }\limits_{x \to 1} {x^2} - \mathop {\lim }\limits_{x \to 1} 1 = {1^2} - 1 = 0\)

\(\mathop {\lim }\limits_{x \to 1} g\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right) = \mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1 = 1 + 1 = 2\)

b) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x} \right) = {1^2} + 1 = 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 + 2 = 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)

c) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - x - 2} \right) = {1^2} - 1 - 2 =  - 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 - 2 =  - 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)

d) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right).g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left[ {\left( {{x^2} - 1} \right)\left( {x + 1} \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^3} + {x^2} - x - 1} \right) = {1^3} + {1^2} - 1 - 1 = 0\\\mathop {\lim }\limits_{x \to 1} f\left( x \right).\mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0.2 = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right).g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right).\mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)

e) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0\\\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}} = \frac{0}{2} = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}}.\end{array}\)

a.

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-ax+2021}-x+1\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(\sqrt{x^2-ax+2021}-x\right)\left(\sqrt{x^2-ax+2021}+x\right)}{\sqrt{x^2-ax+2021}+x}+1\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-ax+2021}{\sqrt{x^2-ax+2021}+x}+1\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x\left(-a+\dfrac{2021}{x}\right)}{x\left(\sqrt{1-\dfrac{a}{x}+\dfrac{2021}{x^2}}+1\right)}+1\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-a+\dfrac{2021}{x}}{\sqrt{1-\dfrac{a}{x}+\dfrac{2021}{x^2}}+1}+1\right)\)

\(=\dfrac{-a+0}{\sqrt{1+0+0}+1}+1=-\dfrac{a}{2}+1\)

\(\Rightarrow a^2=-\dfrac{a}{2}+1\Rightarrow2a^2+a-2=0\)

Pt trên có 2 nghiệm pb nên có 2 giá trị a thỏa mãn

b.

\(\lim\limits_{x\rightarrow-1}f\left(x\right)=\lim\limits_{x\rightarrow-1}\dfrac{x^3+1}{x+1}\)

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

\(=1+1+1=3\)

\(f\left(-1\right)=3a\)

Hàm gián đoạn tại điểm \(x_0=-1\) khi:

\(\lim\limits_{x\rightarrow-1}f\left(x\right)\ne f\left(-1\right)\Rightarrow3\ne3a\)

\(\Rightarrow a\ne1\)

\(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} x = 1\).

\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( { - {x^2}} \right) =  - {1^2} =  - 1\).

Vì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ - }} {\rm{ }}f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).

a) Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì, \({x_n} >  - 1\) và \({x_n} \to  - 1\). Khi đó \(f\left( {{x_n}} \right) = x_n^2 + 2\)

Ta có: \(\lim f\left( {{x_n}} \right) = \lim \left( {x_n^2 + 2} \right) = \lim x_n^2 + \lim 2 = {\left( { - 1} \right)^2} + 2 = 3\)

Vậy \(\mathop {\lim }\limits_{x \to  - {1^ + }} f\left( x \right) = 3\).

Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì, \({x_n} <  - 1\) và \({x_n} \to  - 1\). Khi đó \(f\left( {{x_n}} \right) = 1 - 2{x_n}\).

Ta có: \(\lim f\left( {{x_n}} \right) = \lim \left( {1 - 2{x_n}} \right) = \lim 1 - \lim \left( {2{x_n}} \right) = \lim 1 - 2\lim {x_n} = 1 - 2.\left( { - 1} \right) = 3\)

Vậy \(\mathop {\lim }\limits_{x \to  - {1^ - }} f\left( x \right) = 3\).

b) Vì \(\mathop {\lim }\limits_{x \to  - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {1^ - }} {\rm{ }}f\left( x \right) = 3\) nên \(\mathop {\lim }\limits_{x \to  - 1} f\left( x \right) = 3\).

a) \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} x = 1\)

b) \(f\left( 1 \right) = 1 \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right).\)

a) Với mọi điểm \({x_0} \in \left( {1;2} \right)\), ta có: \(f\left( {{x_0}} \right) = {x_0} + 1\).

\(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = \mathop {\lim }\limits_{x \to {x_0}} \left( {x + 1} \right) = {x_0} + 1\).

Vì \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = f\left( {{x_0}} \right) = {x_0} + 1\) nên hàm số \(y = f\left( x \right)\) liên tục tại mỗi điểm \({x_0} \in \left( {1;2} \right)\).

b) \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 1} \right) = 2 + 1 = 3\).

\(f\left( 2 \right) = 2 + 1 = 3\).

\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = f\left( 2 \right)\).

c) \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x + 1} \right) = 1 + 1 = 2\)

\(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = k \Leftrightarrow 2 = k \Leftrightarrow k = 2\)

Vậy với \(k = 2\) thì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = k\).