a) Ta có \(t = \frac{1}{x},\) nên khi x tiến đến 0 thì t tiến đến dương vô cùng do đó

\(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = \mathop {\lim }\limits_{t \to  + \infty } {\left( {1 + \frac{1}{t}} \right)^t} = e\)

b) \(\ln y = \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \frac{1}{x}\ln \left( {1 + x} \right)\)

\(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\)

c) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)

Đổi biến \(\cos x=y^{20}\). Khi \(x\rightarrow0\) thì \(y\rightarrow0\). Ta có :

\(L=\lim\limits_{y\rightarrow0}\frac{y^5-y^4}{1-y^{40}}=-\lim\limits_{y\rightarrow0}\frac{y^4\left(y-1\right)}{y^{40}-1}\)

    \(=-\lim\limits_{y\rightarrow0}\frac{y-1}{\left(y-1\right)\left(y^{39}+y^{38}+.....+y+1\right)}=-\frac{1}{40}\)

1. Ta có : \(\lim\limits_{x\rightarrow0}\frac{\tan ax}{\tan bx}=\lim\limits_{x\rightarrow0}\left(\frac{\sin ax}{\sin bx}.\frac{\cos ax}{\cos bx}\right)=\lim\limits_{x\rightarrow0}\frac{\sin ax}{\sin bx}=\lim\limits_{x\rightarrow0}\left(\frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}}.\frac{ax}{bx}\right)=\frac{a}{b}\frac{\lim\limits_{x\rightarrow0}\frac{\sin ax}{ax}}{\lim\limits_{x\rightarrow0}\frac{\sin bx}{bx}}=\frac{a}{b}\frac{\lim\limits_{y\rightarrow0}\frac{\sin y}{y}}{\lim\limits_{z\rightarrow0}\frac{\sin z}{z}}=\frac{a}{b}\)

2. Ta có : \(\lim\limits_{x\rightarrow0}\frac{1-\cos ax}{x^2}=\lim\limits_{x\rightarrow0}\frac{2\sin^2\frac{ax}{2}}{x^2}=\lim\limits_{x\rightarrow0}\left[\left(\frac{\sin\frac{ax}{2}.\sin\frac{ax}{2}}{\frac{ax}{2}.\frac{ax}{2}}\right).\frac{a^2}{2}\right]\)

                                   \(=\frac{a^2}{2}\left(\lim\limits_{y\rightarrow0}\frac{\sin y}{y}\right)^2=\frac{a^2}{2}\)

a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)

b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)

\(L=\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan^3x-3\tan x}{\cos\left(x+\frac{\pi}{6}\right)}=\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan x\left(\tan^2x-3\right)}{\cos\left(x+\frac{\pi}{6}\right)}\)

    \(=\sqrt{3}\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\left(\tan x-\sqrt{3}\right)\left(\tan x+\sqrt{3}\right)}{\sin\left(\frac{\pi}{3}-x\right)}=\sqrt{3}.2\sqrt{3}\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan x-\sqrt{3}}{\sin\left(\frac{\pi}{3}-x\right)}\)

    \(=6\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\sin\left(\frac{\pi}{3}-x\right)}{\cos x.\cos\frac{\pi}{3}\sin\left(\frac{\pi}{3}-x\right)}=-12\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{1}{\cos x}=-24\)

a) \(\mathop {\lim }\limits_{x \to  - 2} \left( {{x^2} + 5x - 2} \right) = \mathop {\lim }\limits_{x \to  - 2} {x^2} + \mathop {\lim }\limits_{x \to  - 2} \left( {5x} \right) - \mathop {\lim }\limits_{x \to  - 2} 2\)

\( = \mathop {\lim }\limits_{x \to  - 2} {x^2} + 5\mathop {\lim }\limits_{x \to  - 2} x - \mathop {\lim }\limits_{x \to  - 2} 2 = {\left( { - 2} \right)^2} + 5.\left( { - 2} \right) - 2 =  - 8\)

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

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

Ta có: \(\mathop {\lim }\limits_{x \to {3^ - }} \left( {2x} \right) = 2\mathop {\lim }\limits_{x \to {3^ - }} x = 2.3 = 6;\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{x - 3}} =  - \infty \)

\( \Rightarrow \mathop {\lim }\limits_{x \to {3^ - }} \frac{{2x}}{{x - 3}} =  - \infty \)

b) \(\mathop {\lim }\limits_{x \to  + \infty } \left( {3x - 1} \right) = \mathop {\lim }\limits_{x \to  + \infty } x\left( {3 - \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to  + \infty } x.\mathop {\lim }\limits_{x \to  + \infty } \left( {3 - \frac{1}{x}} \right)\)

Ta có: \(\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty ;\mathop {\lim }\limits_{x \to  + \infty } \left( {3 - \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to  + \infty } 3 - \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{x} = 3 - 0 = 3\)

\( \Rightarrow \mathop {\lim }\limits_{x \to  + \infty } \left( {3x - 1} \right) =  + \infty \)

a: \(\lim\limits_{x\rightarrow-1^+}x+1=0\)

=>\(\lim\limits_{x\rightarrow-1^+}\dfrac{1}{x+1}=+\infty\)

b: \(\lim\limits_{x\rightarrow-\infty}1-x^2=\lim\limits_{x\rightarrow-\infty}\left[x^2\left(\dfrac{1}{x^2}-1\right)\right]\)

\(=-\infty\)

c: \(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=\lim\limits_{x\rightarrow3^-}=\dfrac{-x}{x-3}\)

\(\lim\limits_{x\rightarrow3^-}x-3=0\)

\(\lim\limits_{x\rightarrow3^-}-x=3>0\)

=>\(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=+\infty\)

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

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

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