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

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}+\sqrt{5x+4}-5}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}-2+\sqrt{5x+4}-3}{x-1}\)

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

Đề câu b là \(...\sqrt{90-6x}\) hay \(\sqrt{9-6x}\) vậy em? Hình như cái sau mới có lý

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

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

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

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

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

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

\(=-\infty\)

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

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

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

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

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

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

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

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

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

Giới hạn đã cho hữu hạn khi \(2x^2+ax+b=0\) có nghiệm \(x=1\)

\(\Rightarrow2+a+b=0\Rightarrow b=-a-2\)

Ta được: \(\lim\limits_{x\rightarrow1}\dfrac{2x^2+ax-a-2}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(x+1\right)+a\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

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

\(=\dfrac{4+a}{2}=\dfrac{1}{4}\)

\(\Rightarrow a=-\dfrac{7}{2}\Rightarrow b=\dfrac{3}{2}\)

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

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

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

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

\(=\dfrac{3}{2\cdot4}=\dfrac{3}{8}\)

=>a=3;b=8

=>a²+b=9+8=17