от math10.com » 30 Яну 2014, 11:44
1.[tex]lim_{x\right \infty }\frac{x^3+3x+2}{3x^3-3x+1}=lim_{x\right \infty }\frac{x^3(1+\frac{3}{x^2}+\frac{2}{x^3})}{3x^3(1-\frac{1}{x^2}+\frac{1}{3x^3})}=lim_{x\right \infty }\frac{(1+\frac{1}{x^2}+\frac{2}{x^3})}{3(1-\frac{3}{x^2}+\frac{1}{x^3})}=\frac{(1+0+0)}{3(1-0+0)}=\frac{1}{3}[/tex]
Може и по Лопитал:
[tex]lim_{x\right \infty }\frac{x^3+3x+2}{3x^3-3x+1}=lim_{x\right \infty }\frac{(x^3+3x+2)'''}{(3x^3-3x+1)'''}=\frac{6}{18}=\frac{1}{3}[/tex]
2.[tex]lim_{x\right 0} \frac{sin 2x}{tg 3x}=lim_{x\right 0} \frac{(sin 2x)'}{(tg 3x)'}=lim_{x\right 0} \frac{2cos 2x}{\frac{3}{cos^2 3x}}=lim_{x\right 0} \frac{2cos 2x . cos^2 3x}{3}=\frac{2.1.1}{3}=\frac{2}{3}[/tex]