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:

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

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

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

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:

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

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

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

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

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

\(\Rightarrow P=2\)

Câu này thiếu -1 trên tử rồi :v

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

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

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