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

\(=\lim\limits_{x\rightarrow3}\dfrac{-x-1}{\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}=...\)

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

\(c=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+4x}-2x-1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{-4x^2}{2x+1+\sqrt[]{4x+1}}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{-4}{2x+1+\sqrt[]{4x+1}}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=...\)

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

\(=\lim\limits_{x\rightarrow3}\dfrac{x\left(x-3\right)\left(x+\sqrt{4x-3}\right)}{\left(x-1\right)\left(x-3\right)\left(x+1+\sqrt{5x+1}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{x\left(x+\sqrt{4x-3}\right)}{\left(x-1\right)\left(x+1+\sqrt{5x+1}\right)}=\dfrac{9}{8}\)

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

Câu b bạn coi lại đề, là \(x\rightarrow-1^-\) hay \(x\rightarrow1^-\) (đúng như đề thì ko phải dạng vô định, cứ thay số rồi bấm máy)

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

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

a/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{x-3}=....\)

Từ 2 câu kia lát tui làm, ăn cơm đã :D

\(a=\lim\limits_{x\rightarrow1^+}\dfrac{x^2-x+1}{x^2-1}=\dfrac{1}{0}=+\infty\)

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

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}-\dfrac{1}{4+2\sqrt[3]{8+x}+\sqrt[3]{\left(8+x\right)^2}}\right)=\dfrac{2}{2}-\dfrac{1}{12}=...\)

\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(\sqrt{2x-2}+2\right)}{2\left(x-3\right)\left(\sqrt{x+6}+3\right)}=\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x-2}+2}{2\left(\sqrt{x+6}+3\right)}=\dfrac{2+2}{2\left(3+3\right)}=...\)

\(\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-80}{x-3}\) hữu hạn \(\Rightarrow f\left(3\right)=80\)

Sử dụng hẳng đẳng thức: \(a-b=\dfrac{a^4-b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{\dfrac{f\left(x\right)-80}{\left[\sqrt[4]{f\left(x\right)+1}+3\right]\left[\sqrt[]{f\left(x\right)+1}+9\right]}}{\left(x-3\right)\left(2x-5\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-80}{x-3}.\dfrac{1}{\left[\sqrt[4]{f\left(x\right)+1}+3\right]\left[\sqrt[]{f\left(x\right)+1}+9\right]\left(2x-5\right)}\)

\(=5.\dfrac{1}{\left(\sqrt[4]{80+1}+3\right)\left(\sqrt[]{80+1}+9\right)\left(2.3-5\right)}\)

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

\(b=\lim\limits_{x\rightarrow3}\dfrac{x^2+x-12}{\left(x-3\right)\left(\sqrt{x^2+x}+2\sqrt{3}\right)}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(x+4\right)}{\left(x-3\right)\left(\sqrt{x^2+x}+2\sqrt{3}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{x+4}{\sqrt{x^2+x}+2\sqrt{3}}=\dfrac{7}{\sqrt{12}+2\sqrt{3}}=...\)

\(c=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(x^3-2x^2+4x\right)}{\left(x^2+1\right)\left(x+2\right)}=\lim\limits_{x\rightarrow-2}\dfrac{x^3-2x^2+4x}{x^2+1}=-\dfrac{24}{5}\)

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

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

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

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

\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)

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

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

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!

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

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

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

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

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

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

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

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

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 \(3-x>0;\forall x< 3\)

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