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) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).

b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).

c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).

d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).

em cảm ơn ạ !

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

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

a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)

b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt[]{2x+1}-1\right)+2-\sqrt[3]{x^2+x+8}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2.2x}{\sqrt[]{2x+1}+1}-\dfrac{x\left(x+1\right)}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{2x+1}+1}-\dfrac{x+1}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}\right)\)

\(=\dfrac{23}{12}\)

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

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

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

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-\left(x+1\right)}{2x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x^2+1}-\left(x+1\right)\right)\left(\sqrt{x^2+1}+x+1\right)}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

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

a. \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{x+2}=\dfrac{1}{4}\)

b. \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}\)

Do \(\lim\limits_{x\rightarrow3^-}\left(-x-3\right)=-6< 0\)

\(\lim\limits_{x\rightarrow3^-}\left(3-x\right)=0\) và \(3-x>0;\forall x< 3\)

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

a: \(\lim\limits_{x\rightarrow-2}x^2-7x+4=\left(-2\right)^2-7\cdot\left(-2\right)+4=22\)

b: \(\lim\limits_{x\rightarrow3}\dfrac{x-3}{x^2-9}=\lim\limits_{x\rightarrow3}\dfrac{1}{x+3}=\dfrac{1}{3+3}=\dfrac{1}{6}\)

c: \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-x-8}{3+\sqrt{x+8}}\cdot\dfrac{1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{-1}{3+\sqrt{x+8}}\)

\(=-\dfrac{1}{6}\)

a/ L'Hospital:

 \(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)

b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)