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\(lim_{x\rightarrow1}\frac{x^3+2x-3}{x^2-x}\)   

\(=lim_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}\)   

\(=lim_{x\rightarrow1}\frac{x^2+x+3}{x}\)   

\(=\frac{1^2+1+3}{1}\)   

\(=5\)   

\(lim_{x\rightarrow1}\frac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}\)   

\(=lim_{x\rightarrow1}\frac{\left(2x+2\right)-\left(3x+1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-x+1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-1\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-1}{\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=\frac{-1}{\sqrt{2\cdot1+2}+\sqrt{3\cdot1+1}}\)   

\(=\frac{-1}{2+2}=\frac{-1}{4}\)

https://drive.google.com/file/d/14Q-YI3szy-rePnIHWGD35RKCWiCXCT6k/view?usp=sharing

https://drive.google.com/file/d/1425SNt8hu4qt2y1kIcnhIvcxPfODsY1T/view?usp=sharing

a) \(lim_{x\rightarrow2}\left(\sqrt{x+2}+2018\right)=lim_{x\rightarrow2}\left(\sqrt{2+2}+2018\right)=2020\)

b)\(lim_{x\rightarrow+\infty}\dfrac{3.4^n+2^n}{5.4^n+3^n}=lim_{x\rightarrow+\infty}\dfrac{3+\left(\dfrac{2}{4}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}=\dfrac{3+0}{5+0}=\dfrac{3}{5}\)

c) \(lim_{x\rightarrow-3}\dfrac{x^2+4x+3}{x^2-9}=lim_{x\rightarrow-3}\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=lim_{x\rightarrow-3}\dfrac{x+1}{x-3}=\dfrac{-3+1}{-3-3}=\dfrac{1}{3}\)

a) limx→2(√x+2+2018)=√2+2+2018=2020limx→2(x+2+2018)=2+2+2018=2020.

b) limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35.

limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13.

a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} - 4.2 + 3 =  - 1\)

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 1\)

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x  + 1}} = \frac{1}{{\sqrt 1  + 1}} = \frac{1}{2}\)

a: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{4+\dfrac{3}{x}}{2}=\dfrac{4}{2}=2\)

b: \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}}{3+\dfrac{1}{x}}=0\)

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

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

                                                \( = 3\mathop {\lim }\limits_{x \to  - 1} \left( {{x^2}} \right) - \mathop {\lim }\limits_{x \to  - 1} x + \mathop {\lim }\limits_{x \to  - 1} 2 = 3.{\left( { - 1} \right)^2} - \left( { - 1} \right) + 2 = 6\)

b) \(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \left( {x + 4} \right) = \mathop {\lim }\limits_{x \to 4} x + \mathop {\lim }\limits_{x \to 4} 4 = 4 + 4 = 8\)

c) \(\mathop {\lim }\limits_{x \to 2} \frac{{3 - \sqrt {x + 7} }}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3 - \sqrt {x + 7} } \right)\left( {3 + \sqrt {x + 7} } \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{3^2} - \left( {x + 7} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}}\)

                                         \( = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{3 + \sqrt {x + 7} }}\)

                                         \( = \frac{{\mathop {\lim }\limits_{x \to 2} \left( { - 1} \right)}}{{\mathop {\lim }\limits_{x \to 2} 3 + \sqrt {\mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 7} }} = \frac{{ - 1}}{{3 + \sqrt {2 + 7} }} =  - \frac{1}{6}\)

a) Đặt \(f\left( x \right) = 2{x^2} - x\).

Hàm số \(y = f\left( x \right)\) xác định trên \(\mathbb{R}\).

Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì thỏa mãn \({x_n} \to 3\) khi \(n \to  + \infty \). Ta có:

\(\lim f\left( {{x_n}} \right) = \lim \left( {2x_n^2 - {x_n}} \right) = 2.\lim x_n^2 - \lim {x_n} = {2.3^2} - 3 = 15\).

Vậy \(\mathop {\lim }\limits_{x \to 3} \left( {2{x^2} - x} \right) = 15\).

b) Đặt \(f\left( x \right) = \frac{{{x^2} + 2x + 1}}{{x + 1}}\).

Hàm số \(y = f\left( x \right)\) xác định trên \(\mathbb{R}\).

Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì thỏa mãn \({x_n} \to  - 1\) khi \(n \to  + \infty \). Ta có:

\(\lim f\left( {{x_n}} \right) = \lim \frac{{x_n^2 + 2{x_n} + 1}}{{{x_n} + 1}} = \lim \frac{{{{\left( {{x_n} + 1} \right)}^2}}}{{{x_n} + 1}} = \lim \left( {{x_n} + 1} \right) = \lim {x_n} + 1 =  - 1 + 1 = 0\).

Vậy \(\mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} + 2x + 1}}{{x + 1}} = 0\).

a) Áp dụng giới hạn một bên thường dùng, ta có : \(\mathop {\lim }\limits_{x \to {4^ + }} \frac{1}{{x - 4}} =  + \infty \)

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

Ta có: \(\mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right) =  - \mathop {\lim }\limits_{x \to {2^ + }} x =  - 2;\mathop {\lim }\limits_{x \to {2^ +}} \frac{1}{{x - 2}} =  +\infty \)

\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{2 - x}} =  - \infty \)