Question

Tính các giới hạn sau:

a) $\underset{x\to 3}{\mathop{\lim }}\,\left( x+2 \right);$

b) $\underset{x\to +\infty }{\mathop{\lim }}\,\left( {{x}^{2}}-x+1 \right).$ 

Answer 1

  

Answer 2

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Answer 5

a) limx→3(x+2)=3+2=5

b) limx→+∞(x^2-x+1)=limx→+∞(x^2.(1-1/x+1/x^2)=+∞

( Do limx→+∞(x^2)=+∞

       limx→+∞(1-1/x+1/x^2)=1>0)

Answer 6

                     

Answer 7

  

Answer 8

a, limx→3(x+2) = 3+2=5

b,  limx→+∞(x2−x+1)= limx→+∞(x2−x+1 limx→+∞[x²(1-1/x+1/x²)]

Vì limx→+∞x2=+∞

limx→+∞(1-1/x+1/x²)= 1>0
 

=> limx→+∞(x2−x+1)=+∞

Answer 9

  

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Answer 18

a) limx->3(x + 2) = 5

b) limx->+8 x^2(1-1/x+1/x^2)

limx->+8 x^2= +8

lĩm->+8 1-1/x+1/x^2 = 1

=>limx->+8 x^2(1-1/x+1/x^2)=+8

Answer 19

  

Answer 20

a) limx→3(x+2)= 5

b) limx→+∞(x2−x+1) = +∞

Answer 21

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Answer 30