1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)

2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)

3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)

4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2x+1}{x-1}=3\rightarrow\lim\limits_{x\rightarrow1}\left(f\left(x\right)-2x+1\right)=0\\ \rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=1\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3f\left(x\right)+1}-x-1}{\sqrt{4x+5}-3x-2}=\dfrac{\sqrt{3.1+1}-1-1}{\sqrt{4.1+5}-3.1-2}=0\)

Lời giải:

a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)

\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$

\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)

b.

 \(\lim\limits_{x\to -1+}(3x+2)=-1<0\)

\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$

\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)

c.

\(\lim\limits_{x\to 2-}(x-15)=-17<0\)

\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$

\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)

a) lim= - 1/0 = - vô cùng

d) lim x(x^99-2)+1/ x(x^49-2)+1 =lim (x^99-2)/(x^49-2)=1

\(a=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2-2x-2\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{x^2-2x-2}{x-3}=\dfrac{3}{2}\)

Câu b bạn coi lại đề, là \(x\rightarrow-1^-\) hay \(x\rightarrow1^-\) (đúng như đề thì ko phải dạng vô định, cứ thay số rồi bấm máy)

\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}\)

 \(=\lim\limits_{x\rightarrow3}\dfrac{1}{\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}=\dfrac{1}{2.\left(4+4+4\right)}=...\)

a/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{x-3}=....\)

Từ 2 câu kia lát tui làm, ăn cơm đã :D

a: \(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[3]{x}-x}{x^2-x}\)

\(=\dfrac{\sqrt[3]{-1}-\left(-1\right)}{\left(-1\right)^2-\left(-1\right)}\)

\(=\dfrac{-1+1}{1+1}=\dfrac{0}{2}=0\)

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

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x^3-x^2\right)-\left(x-1\right)}{x^3-x-2x+2}\)

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

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

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

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

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

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)}=\lim\limits_{x\rightarrow1}\dfrac{x+1}{x+2}=\dfrac{1+1}{1+2}=\dfrac{2}{3}\)

\(a=\lim\limits_{x\rightarrow1^+}\dfrac{x^2-x+1}{x^2-1}=\dfrac{1}{0}=+\infty\)

\(b=\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8+x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2x}{\sqrt{1+x}+1}-\dfrac{x}{4+2\sqrt[3]{8+x}+\sqrt[3]{\left(8+x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}-\dfrac{1}{4+2\sqrt[3]{8+x}+\sqrt[3]{\left(8+x\right)^2}}\right)=\dfrac{2}{2}-\dfrac{1}{12}=...\)

\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(\sqrt{2x-2}+2\right)}{2\left(x-3\right)\left(\sqrt{x+6}+3\right)}=\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x-2}+2}{2\left(\sqrt{x+6}+3\right)}=\dfrac{2+2}{2\left(3+3\right)}=...\)

a) Ta có (x - 2)² = 0 và (x - 2)² > 0 với ∀x ≠ 2 và (3x - 5) = 3.2 - 5 = 1 > 0.

Do đó = +∞.

b) Ta có (x - 1) và x - 1 < 0 với ∀x < 1 và (2x - 7) = 2.1 - 7 = -5 <0.

Do đó = +∞.

c) Ta có (x - 1) = 0 và x - 1 > 0 với ∀x > 1 và (2x - 7) = 2.1 - 7 = -5 < 0.

Do đó = -∞.

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}\) hữu hạn nên \(f\left(x\right)-16=0\) có nghiệm \(x=1\)

\(\Rightarrow f\left(1\right)=16\)

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}.\dfrac{1}{\sqrt{2f\left(x\right)+4}+6}=24.\dfrac{1}{\sqrt{2.16+4}+6}=2\)

Tham khảo:

Vì hàm số có giới hạn hữu hạn tại x=1 nên biểu thức tử nhận x=1 làm nghiệm, hay 1+a+b=0.

Áp dụng vào giả thiết, được

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

\(\Leftrightarrow^{lim}_{x\rightarrow1}\dfrac{x+1+a}{x+1}=-\dfrac{1}{2}\Leftrightarrow\dfrac{2+a}{2}=-\dfrac{1}{2}\Leftrightarrow a=-3\)

\(\Rightarrow b=2\)

Lời giải:
Vì $x^2-1\to 0$ khi $x\to 1$ nên để giới hạn đã cho hữu hạn thì $x^2+ax+b$ nhận $x=1$ là nghiệm 

$\Leftrightarrow 1+a+b=0$

$\Leftrightarrow b=-a-1$

Khi đó:
\(\lim\limits_{x\to 1}\frac{x^2+ax+b}{x^2-1}=\lim\limits_{x\to 1}\frac{x^2+ax-a-1}{x^2-1}=\lim\limits_{x\to 1}\frac{(x-1)(x+1+a)}{(x-1)(x+1)}=\lim\limits_{x\to 1}\frac{x+a+1}{x+1}\)

\(=\frac{a+2}{2}=\frac{-1}{2}\Rightarrow a+2=-1\Rightarrow a=-3\)

$b=-a-1=3-1=2$