Question

Tính các giới hạn sau:

a) $\underset{x\to 2}{\mathop{\lim }}\,\left( \sqrt{x+2}+2018 \right)$.

b) $\underset{n\to +\infty }{\mathop{\lim }}\,\dfrac{{{3.4}^{n}}+{{2}^{n}}}{{{5.4}^{n}}+{{3}^{n}}}$.

c) $\underset{x\to -3}{\mathop{\lim }}\,\dfrac{{{x}^{2}}+4x+3}{{{x}^{2}}-9}$. 

Answer 1

  

Answer 2

a) \(lim_{x\rightarrow2}\left(\sqrt{x+2}+2018\right)=lim_{x\rightarrow2}\left(\sqrt{2+2}+2018\right)=2020\)

b)\(lim_{x\rightarrow+\infty}\dfrac{3.4^n+2^n}{5.4^n+3^n}=lim_{x\rightarrow+\infty}\dfrac{3+\left(\dfrac{2}{4}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}=\dfrac{3+0}{5+0}=\dfrac{3}{5}\)

c) \(lim_{x\rightarrow-3}\dfrac{x^2+4x+3}{x^2-9}=lim_{x\rightarrow-3}\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=lim_{x\rightarrow-3}\dfrac{x+1}{x-3}=\dfrac{-3+1}{-3-3}=\dfrac{1}{3}\)

Answer 3

a) limx→2(√x+2+2018)=√2+2+2018=2020limx→2(x+2+2018)=2+2+2018=2020.

 

b) limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35.

 

limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13.

 

Answer 4

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Answer 7

  

 

 

Answer 8

a) limx→2(√x+2+2018)=√2+2+2018=2020limx→2(x+2+2018)=2+2+2018=2020

 

b) limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35

limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35

c) limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13

   

Answer 9

  

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Answer 16

a, =\(\sqrt{2+2}\) +2018=2020
b, =lim_x→+∞\(\dfrac{3+\left(\dfrac{2}{4}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}\)=\(\dfrac{3}{5}\)

c, =lim_x→-3\(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)=lim_x→-3\(\dfrac{x+1}{x-3}\)=\(\dfrac{1}{3}\)

 

Answer 17

  

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Answer 26

a)

\(=\sqrt{2+2}+2018=2020\)

b)

\(\lim\limits_{\rightarrow+\infty}\dfrac{3.4^n+2^n}{5.4^n+3^n}=\lim\limits_{\rightarrow+\infty}\dfrac{3+\dfrac{2^n}{4^n}}{5+\dfrac{3^n}{4^n}}=\lim\limits_{\rightarrow+\infty}\dfrac{3+\left(\dfrac{1}{2}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}=\dfrac{3}{5}\)

Answer 27

  

Answer 28

a) lim_{x->2 ( \(\sqrt{x+2}\)+2018)= \(\sqrt{2+2}+2018=2020\)}

b) lim _x->+∞ \(\dfrac{3.4^n+2^n}{5.4^n+3^n}=lim_{n->+\infty}\dfrac{3+\dfrac{2^n}{4^n}}{5+\dfrac{3^n}{4^n}}=lim_{n->+\infty}\dfrac{3+\left(\dfrac{1}{2}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}=\dfrac{3}{5}\)

c)\(lim_{x->-3}\dfrac{x^2+4x+3}{x^2-9}=lim_{x->-3}\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=lim_{x->-3}\dfrac{x+1}{x-3}=\dfrac{1}{3}\)

Answer 29

  

Answer 30