\(=\lim\limits\dfrac{n^2+an+5-n^2-1}{\sqrt{n^2+an+5}+\sqrt{n^2+1}}=\lim\limits\dfrac{an+4}{\sqrt{n^2+an+5}+\sqrt{n^2+1}}\)

\(=\lim\limits\dfrac{\dfrac{an}{n}+\dfrac{4}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{an}{n^2}+\dfrac{5}{n^2}}+\sqrt{\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}}=\dfrac{a}{1+1}=\dfrac{a}{2}\)

\(\lim\limits\left(u_n\right)=-1\Rightarrow\dfrac{a}{2}=-1\Rightarrow a=-2\)

\(\lim\limits\left(2-3n\right)^4\left(n+1\right)^3=\lim n^7\left(3-\dfrac{2}{n}\right)^4\left(1+\dfrac{1}{n}\right)^3=+\infty\)

\(\lim\left(\sqrt[3]{n+4}-\sqrt[3]{n+1}\right)=\lim\dfrac{3}{\sqrt[3]{\left(n+4\right)^2}+\sqrt[3]{\left(n+4\right)\left(n+1\right)}+\sqrt[3]{\left(n+1\right)^2}}=0\)

\(\lim\left(\sqrt[3]{8n^3+3n^2+4}-2n+6\right)=\lim\dfrac{8n^3+3n^2+4-\left(2n-6\right)^3}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)

\(=\lim\dfrac{75n^2-216n+220}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)

\(=\lim\dfrac{75-\dfrac{216}{n}+\dfrac{220}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}+\dfrac{4}{n^3}\right)^2}+\left(2-\dfrac{6}{n}\right)\sqrt[3]{8+\dfrac{3}{n}+\dfrac{4}{n^3}}+\left(2-\dfrac{6}{n}\right)^2}\)

\(=\dfrac{75}{\sqrt[3]{8^2}+2.\sqrt[3]{8}+2^2}=...\)

d.

\(\lim\left(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\right)\)

\(=\lim\left(\sqrt[3]{8n^3+3n^2-2}-\sqrt[3]{8n^3-5n^2}\right)\)

\(=\lim\dfrac{8n^3+3n^2-2-\left(8n^3-5n^2\right)}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)

\(=\lim\dfrac{8n^2-2}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)

\(=lim\dfrac{8-\dfrac{2}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)^2}+\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)\left(8-\dfrac{5}{n}\right)}+\sqrt[3]{\left(8-\dfrac{5}{n}\right)^2}}\)

\(=\dfrac{8}{\sqrt[3]{8^2}+\sqrt[3]{8.8}+\sqrt[3]{8^2}}=...\)

Hiện tại mới nghĩ được câu b thôi

b/ \(u_1=\dfrac{1}{2};u_2=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3};u_3=\dfrac{1}{2-\dfrac{2}{3}}=\dfrac{3}{4}...\)

Nhận thấy \(u_n=\dfrac{n}{n+1}\) , ta sẽ chứng minh bằng phương pháp quy nạp

\(n=k\Rightarrow u_k=\dfrac{k}{k+1}\)

Chứng minh cũng đúng với \(\forall n=k+1\)

\(\Rightarrow u_{k+1}=\dfrac{k+1}{k+2}\)

Ta có: \(u_{k+1}=\dfrac{1}{2-u_k}=\dfrac{1}{2-\dfrac{k}{k+1}}=\dfrac{k+1}{k+2}\)

Vậy biểu thức đúng với \(\forall n\in N\left(n\ne0\right)\)

\(\Rightarrow limu_n=lim\dfrac{n}{n+1}=lim\dfrac{1}{1+\dfrac{1}{n}}=1\)

\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)

\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)

\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)

\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)

\(u_2=\sqrt{2}\left(2+3\right)-3=5\sqrt{2}-3\)

\(u_3=\sqrt{\dfrac{3}{2}}.5\sqrt{2}-3=5\sqrt{3}-3\)

\(u_4=\sqrt{\dfrac{4}{3}}.5\sqrt{3}-3=5\sqrt{4}-3\)

....

\(\Rightarrow u_n=5\sqrt{n}-3\)

\(\Rightarrow\lim\limits\dfrac{u_n}{\sqrt{n}}=\lim\limits\dfrac{5\sqrt{n}-3}{\sqrt{n}}=5\)

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

\(=\lim\limits_{x\rightarrow0}\dfrac{3x+2}{\sqrt{3x^2+2}+\sqrt{2-2x}}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)