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

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

a. \(lim_{x\rightarrow3}\dfrac{x^3-27}{3x^2-5x-2}=\dfrac{3^3-27}{3.3^2-5.3-2}=\dfrac{0}{10}=0\)

b. \(lim_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{4x^2-3x-2}=\dfrac{\sqrt{2+2}-2}{4.2^2-3.2-2}=\dfrac{0}{8}=0\)

c. \(lim_{x\rightarrow1}\dfrac{1-x^2}{x^2-5x+4}=lim_{x\rightarrow1}\dfrac{\left(1-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-4\right)}=lim_{x\rightarrow1}\dfrac{-\left(x+1\right)}{x-4}=\dfrac{-\left(1+1\right)}{1-4}=\dfrac{2}{3}\)

d. Câu này mình chịu, nhìn đề hơi lạ so với bình thường hehe

1/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7-9\right)\left(2+\sqrt{x+3}\right)}{\left(4-x-3\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(2+\sqrt{x+3}\right)}{\left(x-1\right)\left(-\sqrt{2x+7}-3\right)}=\dfrac{2.4}{-6}=-\dfrac{4}{3}\)

2/ \(=\lim\limits_{x\rightarrow1^-}\dfrac{2.1-3}{1-1}=-\infty\)

3/ \(=\lim\limits_{x\rightarrow2^+}\dfrac{3-x}{x-2}=+\infty\)

4/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-\dfrac{8x^3}{x^2}+\dfrac{9x^2}{x^2}+\dfrac{x}{x^2}-\dfrac{1}{x^2}}{\dfrac{5x^2}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-8x}{5}=\pm\infty\)

5/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}}+\dfrac{2x}{x}-\dfrac{1}{x}}{\dfrac{2x}{x}+\dfrac{7}{x}}=\dfrac{1}{2}\)

1: \(\lim\limits_{x\rightarrow4}\dfrac{1-x}{\left(x-4\right)^2}=-\infty\) 

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow4}1-x=1-4=-3< 0\\\lim\limits_{x\rightarrow4}\left(x-4\right)^2=\left(4-4\right)^2=0\end{matrix}\right.\)

2: \(\lim\limits_{x\rightarrow3^+}\dfrac{2x-1}{x-3}=+\infty\)

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow3^+}2x-1=2\cdot3-1=5>0\\\lim\limits_{x\rightarrow3^+}x-3=3-3>0\end{matrix}\right.\) và x-3>0

3: \(\lim\limits_{x\rightarrow2^+}\dfrac{-2x+1}{x+2}\)

\(=\dfrac{-2\cdot2+1}{2+2}=\dfrac{-3}{4}\)

4: \(\lim\limits_{x\rightarrow1^-}\dfrac{3x-1}{x+1}=\dfrac{3\cdot1-1}{1+1}=\dfrac{2}{2}=1\)