Question

sa se calculeze limita: (clasa 11 fara derivate)

Answer 1

Răspuns:

Explicație pas cu pas:

$\lim_{n \to \infty} \frac{3\cdot 2^n+2\cdot 3^n^+^1}{5\cdot 4^n^-^1-4\cdot 5^n}$

dam factor comun

$\lim_{n \to \infty} \frac{3^n^+^1(\frac{3\cdot2^n}{3\cdot 3^n}+2) }{5^n(\frac{5\cdot 4^n^-^1}{5\cdot5^n^-^1}-4) }$

se simplifica 3 cu 3 si 5 cu 5

$\lim_{n \to \infty} \frac{3^n^+^1(\frac{2^n}{ 3^n}+2) }{5^n(\frac{4^n^-^1}{5^n^-^1}-4) }$

$\lim_{n \to \infty} \frac{3^n^+^1[(\frac{2}{ 3})^n+2] }{5^n[(\frac{4}{5})^n^-^1-4]}$

orice fractie subunitara la puterea ∞ tinde catre 0

adica $(\frac{2}{3} )^n, (\frac{4}{5} )^n^-^1$ tind catre 0

Deci ne ramane

$\lim_{n \to \infty} \frac{3^n\cdot 3\cdot 2}{5^n\cdot(-4)} = \lim_{n \to \infty} (\frac{3}{5} )^n\cdot (\frac{3}{-2} )$=0

pentru ca $(\frac{3}{5} )^n$ tinde catre 0, fiind subunitara

Sper ca ai inteles!