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

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

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

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

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) \(\mathop {\lim }\limits_{x \to  - 3} {x^2};\)            

Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì thỏa mãn \(\lim {x_n} =  - 3.\)

Ta có \(\lim x_n^2 = {\left( { - 3} \right)^2} = 9\)

Vậy \(\mathop {\lim }\limits_{x \to  - 3} {x^2} = 9.\)

b) \(\mathop {\lim }\limits_{x \to 5} \frac{{{x^2} - 25}}{{x - 5}}.\)

Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì thỏa mãn \(\lim {x_n} = 5.\)

Ta có \(\lim \frac{{{x_n}^2 - 25}}{{{x_n} - 5}} = \lim \frac{{\left( {{x_n} - 5} \right)\left( {{x_n} + 5} \right)}}{{{x_n} - 5}} = \lim \left( {{x_n} + 5} \right) = \lim {x_n} + 5 = 5 + 5 = 10\)

Vậy \(\mathop {\lim }\limits_{x \to 5} \frac{{{x^2} - 25}}{{x - 5}} = 10.\)

Ta có : (...) = \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)-\left[\sqrt[3]{x^2+7}-\left(x+1\right)\right]}{x^2-1}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{5-x^3-\left(x+1\right)^2}{\left(\sqrt{5-x^3}+x+1\right)\left(x^2-1\right)}\)  

\(=\lim\limits_{x\rightarrow1}\dfrac{-x^3-x^2-2x+4}{...}\)  \(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)\left(x-1\right)}{...}\)   

= \(\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)}{\left(x+1\right)\left(\sqrt{5-x^3}+x+1\right)}=\dfrac{-7}{8}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{x^2+7-x^3-3x^2-3x-1}{\left(x^2-1\right)\left[\sqrt[3]{\left(x+7\right)^2}+\left(x+1\right)\sqrt[3]{x^2+7}+\left(x+1\right)^2\right]}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)\left(x-1\right)}{...}\)  

\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)}{\left(x+1\right)\left[\sqrt[3]{\left(x^2+7\right)^2}+\sqrt[3]{x^2+7}\left(x+1\right)+\left(x+1\right)^2\right]}\)

\(=\dfrac{-\left(1+3+6\right)}{\left(1+1\right)\left(4+2.2+4\right)}=\dfrac{-5}{12}\)

Suy ra : \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}=\dfrac{-7}{8}+\dfrac{5}{12}=\dfrac{-11}{24}\)

Cách 2 : Tách : \(\sqrt{5-x^3}-2-\left(\sqrt[3]{x^2+7}-2\right)\) -> Dùng liên hợp 

a: \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+10}-4}{3x-9}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2x+10-16}{3x-9}\cdot\dfrac{1}{\sqrt{2x+10}+4}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2\left(x-3\right)}{3\left(x-3\right)\cdot\left(\sqrt{2x+10}+4\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2}{3\left(\sqrt{2x+10}+4\right)}\)

\(=\dfrac{2}{3\cdot\sqrt{6+10}+3\cdot4}=\dfrac{2}{3\cdot4+3\cdot4}=\dfrac{2}{24}=\dfrac{1}{12}\)

b: \(\lim\limits_{x\rightarrow7}\dfrac{\sqrt{4x+8}-6}{x^2-9x+14}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4x+8-36}{\sqrt{4x+8}+6}\cdot\dfrac{1}{\left(x-2\right)\left(x-7\right)}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4x-28}{\left(\sqrt{4x+8}+6\right)\cdot\left(x-2\right)\left(x-7\right)}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4}{\left(\sqrt{4x+8}+6\right)\left(x-2\right)}\)

\(=\dfrac{4}{\left(\sqrt{4\cdot7+8}+6\right)\left(7-2\right)}\)

\(=\dfrac{4}{5\cdot12}=\dfrac{4}{60}=\dfrac{1}{15}\)

c: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{2x^2-9x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{2x^2-10x+x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-5\right)\left(2x+1\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{x-3}{2x+1}=\dfrac{5-3}{2\cdot5+1}=\dfrac{2}{11}\)

\(A=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2017\right)\left(\sqrt[5]{1-5x}-1\right)+x^2}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{5x\left(x^2+2017\right)}{\sqrt[5]{\left(1-5x\right)^4}+\sqrt[5]{\left(1-5x\right)^3}+\sqrt[5]{\left(1-5x\right)^2}+\sqrt[5]{1-5x}+1}+x^2}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(-\dfrac{5\left(x^2+2017\right)}{\sqrt[5]{\left(1-5x\right)^4}+\sqrt[5]{\left(1-5x\right)^3}+\sqrt[5]{\left(1-5x\right)^2}+\sqrt[5]{1-5x}+1}+x\right)\)

\(=-2017\)

dễ thấy hàm số trên có dạng 0/0

áp dụng quy tắc l'Hôpital 

\(A=_{\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2017\right)\sqrt[5]{1-5x}-2017}{x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\left(x^2+2017\right)\sqrt[5]{1-5x}-2017\right)'}{\left(x\right)'}}\)

\(A=\lim\limits_{x\rightarrow0}\dfrac{-x^2-2017}{\sqrt[5]{\left(1-5x\right)^4}}+2x\sqrt[5]{1-5x}=\dfrac{-2017}{1}=-2017\)

Vì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 3 \ne \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 5\) nên không tồn tại giới hạn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)

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

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) 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: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)