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

a: \(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[3]{x}-x}{x^2-x}\)

\(=\dfrac{\sqrt[3]{-1}-\left(-1\right)}{\left(-1\right)^2-\left(-1\right)}\)

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

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

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

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

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

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

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

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

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

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

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

Dạng 0/0 một là phân tích đa thức thành nhân tử để rút gọn mẫu khỏi dạng 0/0. Hoặc là nhân liên hợp

a/ \(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x-\dfrac{3}{2}\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{x}{x^2}-\dfrac{3}{2x^2}}{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}}=0\)

b/ \(=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(2x^2+x+1\right)\left[\left(\sqrt[3]{x+5}\right)^2+2\sqrt[3]{x+5}+4\right]}{x-3}\)

\(=\left(2.3^2+3+1\right)\left[\left(\sqrt[3]{3+5}\right)^2+2\sqrt[3]{3+5}+4\right]=...\)

bn nên đăng ở môn cần nha!

\(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt[3]{2x+12}+x}{x^2+2x}\)

\(=\lim\limits_{x\rightarrow-2}\left(\dfrac{2x+12+x^3}{\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2}\cdot\dfrac{1}{x^2+2x}\right)\)

\(=\lim\limits_{x\rightarrow-2}\left(\dfrac{x^3+2x^2-2x^2-4x+6x+12}{\left(\sqrt[3]{\left(2x+12\right)^2}+x\cdot\sqrt[3]{2x+12}+x^2\right)\cdot x\cdot\left(x+2\right)}\right)\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(x^2-2x+6\right)}{\left(\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2\right)\cdot x\cdot\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2-2x+6}{x\cdot\left(\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2\right)}\)

\(=\dfrac{\left(-2\right)^2-2\cdot\left(-2\right)+6}{\sqrt[3]{\left(-2\cdot2+12\right)^2}-\left(-2\right)\cdot\sqrt[3]{2\cdot\left(-2\right)+12}+\left(-2\right)^2}\)

\(=\dfrac{4+4+6}{\sqrt[3]{64}+2\cdot\sqrt[3]{8}+4}\)

\(=\dfrac{14}{8+2\cdot2+4}=\dfrac{14}{12+4}=\dfrac{14}{16}=\dfrac{7}{8}\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{x+2}}{x^3-8}\)

\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x^2-x-2}{x+\sqrt{x+2}}\cdot\dfrac{1}{x^3-8}\right)\)

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

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

\(=\dfrac{2+1}{\left(2+\sqrt{2+2}\right)\left(2^2+2\cdot2+4\right)}\)

\(=\dfrac{3}{\left(2+2\right)\left(4+4+4\right)}=\dfrac{3}{12\cdot4}=\dfrac{1}{4\cdot4}=\dfrac{1}{16}\)

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

Hiển nhiên là cách đầu sai rồi em

Khi đến \(\lim x^2\left(1-1\right)=+\infty.0\) là 1 dạng vô định khác, đâu thể kết luận nó bằng 0 được