Xét 2 khai triển:

\(\left(x+1\right)^{2018}=C_{2018}^0+C_{2018}^1x+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}\)

\(\left(x-1\right)^{2018}=C_{2018}^0-C_{2018}^1x+C_{2018}^2x^2-...+C_{2018}^{2018}x^{2018}\)

Cộng vế với vế:

\(\left(x+1\right)^{2018}+\left(x-1\right)^{2018}=2\left(C_{2018}^0+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}\right)\)

\(\Leftrightarrow C_{2018}^0+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}=\frac{1}{2}\left(x+1\right)^{2018}+\frac{1}{2}\left(x-1\right)^{2018}\)

\(\Rightarrow\lim\limits_{x\rightarrow1}=\frac{\frac{1}{2}\left(x+1\right)^{2018}+\frac{1}{2}\left(x-1\right)^{2018}-2^{2017}}{x-1}=\lim\limits_{x\rightarrow1}\frac{1009\left(x+1\right)^{2017}+1009\left(x-1\right)^{2017}}{1}=1009.2^{2017}\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{-a\sqrt{1+\dfrac{1}{x^2}}+\dfrac{2017}{x}}{1+\dfrac{2018}{x}}=-a\Rightarrow a=-\dfrac{1}{2}\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{bx+1}{\sqrt{x^2+bx+1}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{b+\dfrac{1}{x}}{\sqrt{1+\dfrac{b}{x}+\dfrac{1}{x^2}}+1}=\dfrac{b}{2}=2\Rightarrow b=4\)

\(\Rightarrow P=2\)

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:

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

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

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

a) \(lim_{x\rightarrow2}\left(\sqrt{x+2}+2018\right)=lim_{x\rightarrow2}\left(\sqrt{2+2}+2018\right)=2020\)

b)\(lim_{x\rightarrow+\infty}\dfrac{3.4^n+2^n}{5.4^n+3^n}=lim_{x\rightarrow+\infty}\dfrac{3+\left(\dfrac{2}{4}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}=\dfrac{3+0}{5+0}=\dfrac{3}{5}\)

c) \(lim_{x\rightarrow-3}\dfrac{x^2+4x+3}{x^2-9}=lim_{x\rightarrow-3}\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=lim_{x\rightarrow-3}\dfrac{x+1}{x-3}=\dfrac{-3+1}{-3-3}=\dfrac{1}{3}\)

a) limx→2(√x+2+2018)=√2+2+2018=2020limx→2(x+2+2018)=2+2+2018=2020.

b) limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35.

limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13.

Ta thấy nó có dạng \(\frac{0}{0}\)

Áp dụng Lopitan ta được

\(lim\frac{\sqrt[2018]{11x+1}-1}{x}=lim\frac{11}{2018\sqrt[2018]{\left(11x+1\right)^{2017}}}=\frac{11}{2018}\)

Câu này thiếu -1 trên tử rồi :v

Tham khảo câu trả lời của mod Lâm  Đọc bị lú rồi :D

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:

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

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

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

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

cho to sua a=1 nhe

\(\lim\limits_{x\rightarrow+\infty}\left(ax-\sqrt{bx^2-2x+2018}\right)=\lim\limits_{x\rightarrow+\infty}x.\lim\limits_{x\rightarrow+\infty}\left(a-\sqrt{b}\right)=\pm\infty\)

Còn tuỳ vào độ lớn của a và b