Question

помогите решить а), б) и в)!!!!!!!!!!!!!

Answer 1

a)$displaystyle lim_{x to infty} frac{3x^2+5x+4}{2x^2-x+1}= lim_{x to infty} frac{(3x^2+5x+4):x^2}{(2x^2-x+1):x^2}=\lim_{x to infty} frac{3+frac{5}{x}+frac{4}{x^2}}{2-frac{1}{x}+frac{1}{x^2}}=frac{3}{2};$

б)$displaystyle lim_{x to -2} frac{x^2-4}{sqrt{1-4x}-3}= lim_{x to -2} frac{(x^2-4)(sqrt{1-4x}+3)}{(sqrt{1-4x}-3)(sqrt{1-4x}+3)}=\lim_{x to -2} frac{(x^2-4)(sqrt{1-4x}+3)}{1-4x-9}=lim_{x to -2} frac{(x-2)(x+2)(sqrt{1-4x}+3)}{-4(x+2)}=\lim_{x to -2} frac{(x-2)(sqrt{1-4x}+3)}{-4}=frac{(-2-2)(sqrt{1+8}+3)}{-4}=6;$

в) $displaystyle tg3x=frac{3tgx-tg^3x}{1-3tg^2x}; sin4x=2sin2xcos2x=4sinxcosx(cos^2x-sin^2x)\lim_{x to 0} frac{tg3x}{sin4x}=lim_{x to 0} frac{frac{3tgx-tg^3x}{1-3tg^2x}}{4sinxcosx(cos^2x-sin^2x)}=\lim_{x to 0} frac{3tgx-tg^3x}{(1-3tg^2x)4sinxcosx(cos^2x-sin^2x)}=\lim_{x to 0} frac{3tgx-tg^3x}{4(sinxcosx-3tgx*sin^2x)(cos^2x-sin^2x)}=$

$displaystyle lim_{x to 0} frac{frac{3sinxcos^2x-sin^3x}{cos^3x}}{4(frac{sinxcos^2x-3sin^3x}{cosx})(cos^2x-sin^2x)}=\lim_{x to 0} frac{3sinxcos^2x-sin^3x}{4cos^3xsinx(frac{cos^2x-3sin^2x}{cosx})(cos^2x-sin^2x)}=\lim_{x to 0} frac{3cos^2x-sin^2x}{4cos^2x(cos^2x-3sin^2x)(cos^2x-sin^2x)}=\frac{3*1-0}{4*1(1-3*0)(1-0)}=frac{3}{4}$