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25 tháng 4 2017

a/ \(\lim\limits_{x\to 1} f(x)=\frac{x^{2}-5x + 6}{x-2} \)

\(<=>\lim\limits_{x\to 1} f(x)=\dfrac{(x-3)(x-2)}{x-2} \)

<=>\(\lim\limits_{x\to 1} f(x)=x-3 \)

\(<=>\lim\limits_{x\to 1} f(x)=-2\)

28 tháng 4 2017

Tôi chẳng thể hiểu nổi

AH
Akai Haruma
Giáo viên
22 tháng 2 2023

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 $x<2$

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

 

 

 

9 tháng 2 2021

a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)

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

1: \(A=\dfrac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)

\(=\dfrac{x^2-xa-x+a}{\left(x-a\right)\left(x^2+ax+a^2\right)}\)

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

\(lim_{x->a}A=lim_{x->a}\left(\dfrac{x-1}{x^2+ax+a^2}\right)\)

\(=\dfrac{a-1}{a^2+a^2+a^2}=\dfrac{a-1}{3a^2}\)

2: \(B=\dfrac{1}{1-x}-\dfrac{3}{1-x^3}\)

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

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

\(lim_{x->1}\left(B\right)=\dfrac{-1-2}{1^2+1+1}=\dfrac{-3}{3}=-1\)

3: \(C=\dfrac{\left(x+h\right)^3-x^3}{h}=\dfrac{\left(x+h-x\right)\left(x^2+2xh+h^2+x^2+hx+x^2\right)}{h}\)

\(=3x^2+3hx\)

\(lim_{h->0}\left(C\right)=3x^2+3\cdot0\cdot x=3x^2\)

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

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

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

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

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

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

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

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