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

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

Ta có:

Chọn C.

\(\lim\limits_{x\rightarrow a}\dfrac{x^4-a^4}{x^2-a^2}=\lim\limits_{x\rightarrow a}\left(x^2+a^2\right)=2a^2\)

\(=\lim\limits_{x\rightarrow\dfrac{1}{2}^-}\dfrac{4.\left(\dfrac{1}{2}\right)+1}{0}=+\infty\)

1: \(\lim\limits_{x\rightarrow4}\dfrac{1-x}{\left(x-4\right)^2}=-\infty\) 

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow4}1-x=1-4=-3< 0\\\lim\limits_{x\rightarrow4}\left(x-4\right)^2=\left(4-4\right)^2=0\end{matrix}\right.\)

2: \(\lim\limits_{x\rightarrow3^+}\dfrac{2x-1}{x-3}=+\infty\)

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow3^+}2x-1=2\cdot3-1=5>0\\\lim\limits_{x\rightarrow3^+}x-3=3-3>0\end{matrix}\right.\) và x-3>0

3: \(\lim\limits_{x\rightarrow2^+}\dfrac{-2x+1}{x+2}\)

\(=\dfrac{-2\cdot2+1}{2+2}=\dfrac{-3}{4}\)

4: \(\lim\limits_{x\rightarrow1^-}\dfrac{3x-1}{x+1}=\dfrac{3\cdot1-1}{1+1}=\dfrac{2}{2}=1\)

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

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

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

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+9}-3+\sqrt{x+16}-4}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x}{\sqrt{x+9}+3}+\dfrac{x}{\sqrt{x+16}+4}}{x}\)

\(\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt{x+9}+3}+\dfrac{1}{\sqrt{x+16}+4}\right)=\dfrac{7}{24}\)

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

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

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