Lời giải:
\(\frac{(x^2+x+1)^{2018}+(x+2)^{2018}-2.3^{2018}}{(x-1)(x+2017)}=\frac{(x^2+x+1)^{2018}-3^{2018}+(x+2)^{2018}-3^{2018}}{(x-1)(x+2017)}\)

\(=\frac{(x^2+x-2)[(x^2+x+1)^{2017}+...+3^{2017}]+(x-1)[(x+2)^{2017}+...+3^{2017}]}{(x-1)(x+2017)}\)

\(=\frac{(x+2)[(x^2+x+1)^{2017}+...+3^{2017}]+(x+2)^{2017}+...+3^{2017}}{x+2017}\)

Do đó:

\(\lim_{x\to 1}\frac{(x^2+x+1)^{2018}+(x+2)^{2018}-2.3^{2018}}{(x-1)(x+2017)}=\lim_{x\to 1}\frac{(x+2)[(x^2+x+1)^{2017}+...+3^{2017}]+(x+2)^{2017}+...+3^{2017}}{x+2017}\)

\(=\frac{3\underbrace{(3^{2017}+3^{2017}+...+3^{2017})}_{2018}+\underbrace{3^{2017}+...+3^{2017}}_{2018}}{1+2017}\)

\(=\frac{3.2018.3^{2017}+2018.3^{2017}}{2018}=3^{2018}+3^{2017}=3^{2017}.4\)

\(\lim\limits_{x\rightarrow1}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}=\lim\limits_{x\rightarrow1}\frac{2016x^{2015}+1}{\frac{1009}{\sqrt{2018x+1}}-\frac{1}{2\sqrt{x+2018}}}=\frac{2017}{\frac{1009}{\sqrt{2019}}-\frac{1}{2\sqrt{2019}}}=2\sqrt{2019}\)

Để hàm liên tục tại \(x=1\)

\(\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Rightarrow k=2\sqrt{2019}\)

2.

\(\lim\limits_{x\rightarrow1}\frac{x^2+ax+b}{x^2-1}=\frac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}a+b+1=0\\\lim\limits_{x\rightarrow1}\frac{2x+a}{2x}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-1\\\frac{a+2}{2}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=0\end{matrix}\right.\) \(\Rightarrow S=1\)

3.

\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{7\left(x-1\right)}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}}{\sqrt{2}\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{7}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}\right)\)

\(=\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{7}{12}\right)=\frac{\sqrt{2}}{12}\)

\(\Rightarrow a+b+c=1+12+0=13\)

Cái \(\sqrt[3]{2.3x+a}\) đúng hay sai đấy bạn? Bạn có gõ nhầm 1 thành a ko?

Sửa đề:

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1.2x+1}\sqrt[3]{2.3x+1}...\sqrt[2018]{2017.2018x+1}-1}{x}\)

Do gõ \(x\rightarrow0\) dưới lim rất tốn thời gian nên mình bỏ qua, bạn tự hiểu tất cả các giới hạn bên dưới đều là \(x\rightarrow0\)

Trước hết ta dùng L'Hopital để tính giới hạn dạng tổng quát sau:

\(lim\dfrac{\sqrt[n]{\left(n-1\right)n.x+1}-1}{x}=lim\dfrac{\left[\left(n-1\right)nx+1\right]^{\dfrac{1}{n}}-1}{x}\)

\(=lim\dfrac{\dfrac{1}{n}\left[\left(n-1\right)nx+1\right]^{\dfrac{1}{n}-1}.\left(n-1\right)n}{x}=n-1\)

Và \(\sqrt{2.3x+1}...\sqrt[n]{\left(n-1\right)n.x+1}=1\) khi \(x=1\)

\(\Rightarrow lim\dfrac{\sqrt[k]{\left(k-1\right)kx+1}...\sqrt[m]{\left(m-1\right)mx+1}\left(\sqrt[n]{\left(n-1\right)nx+1}-1\right)}{x}=n-1\)

với mọi \(m;k\) (vì đằng nào cái cụm nhân đằng trước cũng ra 1, ko ảnh hưởng)

Áp dụng vào bài toán:

\(lim\dfrac{\sqrt{1.2x+1}\sqrt[3]{2.3x+1}...\sqrt[2018]{2017.2018x+1}-1}{x}\)

\(=lim\dfrac{\sqrt[3]{2.3x+1}...\sqrt[2018]{2017.2018x+1}\left(\sqrt{2.3x+1}-1\right)}{x}+\) \(lim\dfrac{\sqrt[4]{3.4x+1}...\sqrt[2018]{2017.2018x+1}\left(\sqrt[3]{2.3x+1}-1\right)}{x}+...\)

\(+lim\dfrac{\sqrt[2018]{2017.2018x+1}-1}{x}\)

\(=2+3+...2017=\dfrac{2016.2019}{2}=2035152\)