lim(x стремится к бесконечности)=(n^2+7)/(4n^2-1)если можете объясните пожалуйста как

lim(x стремится к бесконечности)=(n^2+7)/(4n^2-1)
если можете объясните пожалуйста как решать.
- Предмет:
  Математика
- Автор:
  tamara
- 6 лет назад

Ответы 3

подскажите плз еще 1 задачу...
- Автор:
  dilly dally4ppv
- 6 лет назад
- 0
lim(x стремится к бесконечности) (5n+3)/n+1
- Автор:
  carolinaygpb
- 6 лет назад
- 0
Разделим дробь на n² (1/n² = 0) $\lim_{n \to \infty} \frac{n^2+7}{4n^2-1} = \lim_{n \to \infty} \frac{1+ \frac{7}{n^2} }{4- \frac{1}{n^2} } = \frac{1+0}{4-0} = \frac{1}{4} = 0.25$
- Автор:
  ibarra
- 6 лет назад
- 0
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lim(x стремится к бесконечности)=(n^2+7)/(4n^2-1)если можете объясните пожалуйста как решать.

Ответы 3

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lim(x стремится к бесконечности)=(n^2+7)/(4n^2-1)
если можете объясните пожалуйста как решать.