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

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