Question

Tính các giới hạn sau:

a. $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{{{x}^{3}}+2x-3}{{{x}^{2}}-x}$;

b. $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}$.

Answer 1

o

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Answer 5

a, =\(lim_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}=lim_{x\rightarrow1}\dfrac{x^2+x+3}{x}=5\)
b,=\(lim_{x\rightarrow1}\dfrac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}=lim_{x\rightarrow1}\dfrac{-1}{\sqrt{2x+2}+\sqrt{3x+1}}=\dfrac{-1}{4}\)

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A. 8

B.0

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Answer 13

a. limx→1x3+2x−3x2−x=limx→1(x−1)(x2+x+3)x(x−1)=limx→1x2+x+3x=5limx→1x3+2x−3x2−x=limx→1(x−1)(x2+x+3)x(x−1)=limx→1x2+x+3x=5.

b. limx→1√2x+2−√3x+1x−1=limx→11−x(x−1)(√2x+2+√3x+1)=limx→1−1√2x+2+√3x+1=−14limx→12x+2−3x+1x−1=limx→11−x(x−1)(2x+2+3x+1)=limx→1−12x+2+3x+1=−14.

Answer 14

a) \(\lim\limits_{x\rightarrow1}\dfrac{x^3+2x-3}{x^2-x}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}=\lim\limits_{x\rightarrow1}\dfrac{x^2+x+3}{x}=\dfrac{1+1+3}{1}=5\)

b)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}=-\dfrac{1}{4}\)

Answer 15

keosi

Answer 16

a) \(\overset{lim}{x\rightarrow1}\dfrac{x^3+2x-3}{x^2-x}=\overset{lim}{x\rightarrow1}\dfrac{\left(x^2+x+3\right)\left(x-1\right)}{x\left(x-1\right)}=\overset{lim}{x\rightarrow1}\dfrac{x^2+x+3}{x}=\dfrac{1^2+1+3}{1}=5\)

b)\(\overset{lim}{x\rightarrow1}\dfrac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}=\overset{lim}{x\rightarrow1}\dfrac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)​ 

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

                                               = \(\overset{lim}{x\rightarrow1}\dfrac{-1}{\sqrt{2x+2}+\sqrt{3x+1}}\)

                                              = \(\dfrac{-1}{\sqrt{2.1+2}+\sqrt{3.1+1}}=-\dfrac{1}{4}\)

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A,=5    B,=0

 

 

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Answer 25

x3+2x−3x2−x=(x−1)(x2+x+3)x(x−1)=x2+x+3x=5x→1lim​x2−xx3+2x−3​=x→1lim​x(x−1)(x−1)(x2+x+3)​=x→1lim​xx2+x+3​=5.

 

b. 2x+2−3x+1x−1=1−x(x−1)(2x+2+3x+1)=−12x+2+3x+1=−14x→1lim​x−12x+2​−3x+1​​=x→1lim​(x−1)(2x+2​+3x+1​)1−x​=x→1lim​2x+2​+3x+1​−1​=−41​.

 

x3+2x−3x2−x=(x−1)(x2+x+3)x(x−1)=x2+x+3x=5x→1lim​x2−xx3+2x−3​=

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