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01100000
@01100000
July 2022
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Решите,пожалуйста.
1.
a)[tex] \lim_{n \to \infty} \frac{2n+1}{5-3n } [/tex]
б)[tex] \lim_{n \to \infty} \frac{2n^{2}-1 }{ n^{2}+5 } [/tex]
в)[tex] \lim_{n \to \infty} \frac{ n^{2} }{2 n^{2}-1 } [/tex]
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amin07am
Verified answer
Ответ ответ ответ ответ ответ ответ ответ
2 votes
Thanks 2
01100000
спасибо большое!
sedinalana
Verified answer
1
limn(2+1/n)/n(5/n-3)=lim(2+1/n)/(5/n-3)=(2+0)/(0-3)=-2/3
x→∞
2
limn²(2-1/n²)/n²(1+5/n²)=lim(2-1/n²)/(1+5/n²)=(2-0)/(1+0)=2
x→∞
3
limnπ/nπ(2-1/π²)=lim1/(2-1/n²)=1/(2-0)=1/2
x→∞
0 votes
Thanks 1
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Answers & Comments
Verified answer
Ответ ответ ответ ответ ответ ответ ответVerified answer
1limn(2+1/n)/n(5/n-3)=lim(2+1/n)/(5/n-3)=(2+0)/(0-3)=-2/3
x→∞
2
limn²(2-1/n²)/n²(1+5/n²)=lim(2-1/n²)/(1+5/n²)=(2-0)/(1+0)=2
x→∞
3
limnπ/nπ(2-1/π²)=lim1/(2-1/n²)=1/(2-0)=1/2
x→∞