\(\Rightarrow1+a+b=0\Leftrightarrow b=-a-1\)

\(\lim\limits_{x\rightarrow1}\dfrac{x^2+ax-a-1}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x+1+a\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{x+1+a}{x+1}=\dfrac{1+1+a}{1+1}=\dfrac{1}{2}\)

\(\Rightarrow a=-1\Rightarrow b=0\)

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

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.

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

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

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

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

Bài 1:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)

Bài 2:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)

\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)

GIới hạn đã cho hữu hạn

\(\Rightarrow\sqrt[3]{13x^2+2x+5}-\sqrt[3]{81x^2+ax+1}=0\) có nghiệm \(x=-1\)

\(\Rightarrow a=18\)

Khi đó:

\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{13x^2+2x+5}-\sqrt[3]{81x^2+18x+1}}{\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt[]{13x^2+2x+5}-\left(1-3x\right)\right)+\left(1-3x-\sqrt[3]{81x^3+18x+1}\right)}{\left(x+1\right)^2}\)

\(=...=\dfrac{17}{16}\)

Vậy nó ko phải dạng vô định, cứ thay số trực tiếp

\(=\frac{2}{0}=+\infty\)

Nếu là mũ 3 thì nó là dạng 0/0 rút gọn được. Nên chắc là đề ghi nhầm đấy

Sry mình ko nhớ 1 chữ về vật lý luôn :<

\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)

\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)

\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)

\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)

\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)

\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)

Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital

Lời giải:

\(\lim\limits _{x\to 0}\frac{(x+a)^3-a^3}{x}=\lim\limits _{x\to 0}\frac{x[(x+a)^2+a(x+a)+a^2]}{x}=\lim\limits _{x\to 0}[(x+a)^2+a(x+a)+a^2]\)

\(=3a^2\)

Để \(\lim\limits _{x\to 0}\frac{(x+a)^3-a^3}{x}=a\) \(\Leftrightarrow 3a^2=a\)

\(\Leftrightarrow 3a^2-a=0\Leftrightarrow a=0; a=\frac{1}{3}\) (có 2 giá trị thực của a)

Đáp án A.

\(4x^3-3x+1=\left(2x-1\right)^2\left(x+1\right)\) có nghiệm kép \(x=\dfrac{1}{2}\)

\(\Rightarrow\sqrt{1+ax^2}-bx-2=0\) có nhiều hơn 1 nghiệm \(x=\dfrac{1}{2}\)

\(\Rightarrow\sqrt{1+\dfrac{a}{4}}=\dfrac{b}{2}+2\Rightarrow\sqrt{a+4}=b+4\) (\(b\ge-4\))

\(\Rightarrow a=b^2+8b+12\)

\(\Rightarrow\sqrt{1+\left(b^2+8b+12\right)x^2}=bx+2\)

\(\Rightarrow1+\left(b^2+8b+12\right)x^2=b^2x^2+4bx+4\)

\(\Rightarrow\left(8b+12\right)x^2-4bx-3=0\)

\(\Rightarrow\left(2x-1\right)\left[\left(4b+6\right)x+3\right]=0\)

\(\Rightarrow\left(4b+6\right)x+3=0\) có nghiệm \(x=\dfrac{1}{2}\)

\(\Rightarrow2b+3+3=0\Rightarrow b=-3\) \(\Rightarrow a=-3\)

Khi đó:

\(\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12\left(2x-1\right)^2}{\left(x+1\right)\left(2x-1\right)^2\left(\sqrt{1-3x^2}+2-3x\right)}\)

\(=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12}{\left(x+1\right)\left(\sqrt{1-3x^2}+2-3x\right)}=-8\)

\(\Rightarrow c=-8\)

Lời giải:

\(\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{(x+1)(2x-1)^2}\)

Để giới hạn hàm đã cho hữu hạn thì $f(x)=\sqrt{1+ax^2}-bx-2$ có nhân tử là $(2x-1)^2$

$f(x)$ có nhân tử $2x-1 \Leftrightarrow f(\frac{1}{2})=0\Leftrightarrow b=\sqrt{4+a}-4$

Khi đó:

$\sqrt{1+ax^2}-bx-2=(2x-1)(2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}})$

Giờ ta cần xác định $a,b$ để $2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}}=0$ với $x=\frac{1}{2}$

$\Leftrightarrow \sqrt{4+a}=1\Leftrightarrow a=-3$

$b=\sqrt{4+a}-4=-3$

\(\lim\limits_{x\to 0,5}\frac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{-3(2x-1)^2(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(2x-1)^2(x+1)}\)

\(=\lim\limits_{x\to 0,5}\frac{-3(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(x+1)}=-2=c\)