Question

Va rog frumos am nevoie

Answer 1

Explicație pas cu pas:

$\lim_{x \rightarrow + \infty } \dfrac{6 \ln + 4 \ln(x^{3} + 3) }{9 \ln + 5 \ln(x^{3} + 3)} =$

$= \lim_{x \rightarrow + \infty } \dfrac{6 \cdot \frac{ \ln x}{\ln(x^{3} + 3)} + 4}{9 \cdot \frac{ \ln x}{\ln(x^{3} + 3)} + 5}$

$= \lim_{x \rightarrow + \infty } \dfrac{6 \cdot \frac{ \frac{1}{x} }{ \frac{3 {x}^{2} }{x^{3} + 3} } + 4}{9 \cdot \frac{ \frac{1}{x} }{ \frac{3 {x}^{2} }{x^{3} + 3} } + 5}$

$= \lim_{x \rightarrow + \infty } \dfrac{6 \cdot \frac{x^{3} + 3}{3 {x}^{3}} + 4}{9 \cdot \frac{x^{3} + 3}{3 {x}^{3}} + 5}$

$= \lim_{x \rightarrow + \infty } \dfrac{6 \cdot \Big ( \frac{1}{3} + \frac{1}{{x}^{3}}\Big) + 4}{9 \cdot \Big ( \frac{1}{3} + \frac{1}{{x}^{3}}\Big) + 5}$

$= \dfrac{6 \cdot \frac{1}{3} + 4}{9 \cdot \frac{1}{3} + 5} = \dfrac{2 + 4}{3 + 5} = \dfrac{6}{8} = \bf \dfrac{3}{4}$