Question Number 66434 by iklima_0412 last updated on 15/Aug/19
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{x}}{\mathrm{1}−\mathrm{2x}} \\ $$
Commented by mathmax by abdo last updated on 15/Aug/19
$${let}\:{f}\left({x}\right)=\frac{{cos}^{\mathrm{2}} {x}−{x}}{\mathrm{1}−\mathrm{2}{x}}\:\Rightarrow\:{for}\:{x}\:\neq\mathrm{0}\:\:{we}\:{have}\:{f}\left({x}\right)=\frac{{x}−{cos}^{\mathrm{2}} {x}}{\mathrm{2}{x}−\mathrm{1}} \\ $$$$=\frac{{x}\left(\mathrm{1}−\frac{{cos}^{\mathrm{2}} {x}}{{x}}\right)}{{x}\left(\mathrm{2}−\frac{\mathrm{1}}{{x}}\right)}\:=\frac{\mathrm{1}−\frac{{cos}^{\mathrm{2}} {x}}{{x}}}{\mathrm{2}−\frac{\mathrm{1}}{{x}}}\:\:{but}\:{lim}_{{x}\rightarrow+\infty} \:\:\:\frac{{cos}^{\mathrm{2}} {x}}{{x}}\:=\mathrm{0}\:{becsuse}\:\mid{cosx}\mid\leqslant\mathrm{1} \\ $$$${lim}_{{x}\rightarrow+\infty} \:\frac{\mathrm{1}}{{x}}\:=\mathrm{0}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by kaivan.ahmadi last updated on 15/Aug/19
$$\equiv{lim}_{{x}\rightarrow\infty} \:\frac{−{x}}{−\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$