Salut,
Atunci când f(x) tinde la 0, atunci:
[tex]\displaystyle\lim_{x\to0}\dfrac{sin(f(x))}{f(x)}=1.\ Limita\ din\ enun\c{t}\ se\ rezolv\breve{a}\ astfel:\\\\\lim_{\substack{x\to0\\x < 0}}\dfrac{sin^2(2x)}{x^2}=\lim_{\substack{x\to0\\x < 0}}\dfrac{4\cdot sin^2(2x)}{4\cdot x^2}=4\cdot\lim_{\substack{x\to0\\x < 0}}\left(\dfrac{sin(2x)}{2x}\right)^2=4\cdot 1^2=4.[/tex]
A fost greu ?
Green eyes.
