Question

limitele dracului,help please

Answer 1

   
[tex]\displaystyle\\\text{Folosim formula lui L'OPITAL care suna asa:} \\ \\ \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)} \\ \\ 1) \\ \lim_{x \to (-6)} \frac{x^2+4x-12}{x^2+2x-24} = \lim_{x \to (-6)} \frac{(x^2+4x-12)'}{(x^2+2x-24)'} = \\ \\ = \lim_{x \to (-6)} \frac{2x+4}{2x+2} =\frac{2\cdot (-6)+4}{2\cdot (-6)+2} = \frac{-12+4}{-12+2} = \frac{-8}{-10}= \frac{8}{10}= \boxed{\frac{4}{5}} [/tex]


[tex]\displaystyle \\ 2) \\ \lim_{x \to 2} \frac{3x^2-10x+8}{2x^2-5x+2} = \lim_{x \to 2} \frac{(3x^2-10x+8)'}{(2x^2-5x+2)'} = \\ \\ = \lim_{x \to 2} \frac{6x-10}{4x-5} =\frac{6\cdot 2-10}{4\cdot 2-5}= \frac{12-10}{8-5}=\boxed{\frac{2}{3}} [/tex]


[tex]\displaystyle \\ 3) \\ \lim_{x \to (-4)} \frac{3x^2+10x-8}{64+x^3} = \lim_{x \to (-4)} \frac{(3x^2+10x-8)'}{(64+x^3)'} = \\ \\ = \lim_{x \to (-4)} \frac{6x+10}{3x^2} =\frac{6\cdot (-4)+10}{3\cdot (-4)^2} =\frac{-24+10}{3\cdot 16} =\frac{-14}{48} =\boxed{-\frac{7}{24}} [/tex]

Answer 2

lim(x→-6) {[x²+4x-12]/[x²+2x-24]}=lim(x→-6) {[(x-2)(x+6)] / [(x+6)(x-4)]}=
=lim(x→-6) {(x-2)/(x-4)}= [-6-2]/[-6-4]=(-8)/(-10)=4/5 

lim(x→2) {[3x²-10x+8]/[2x²-5x+2]}=lim(x→2) {[(x-2)(3x-4)]/[(x-2)(2x-1)]}=
=lim(x→2) {(3x-4)/(2x-1)}=(3·2-4)/(2·2-1)=2/3

lim(x→-4) {[3x²+10x-8]/[64+x³]}=lim(x→-4) {[(x+4)(3x-2)]/[(x+4)(x²-4x+16)]}
=lim(x→-4) {[3x-2]/{[x²-4x+16]}=[3·(-4)-2]/[(-4)²-4·(-4)+16)=[-14]/48=-(7/24)

Aici despartim (ecuatiile de gradul 2) in (2 ecuatii de gradul 1), asa ca in sus si in jos sa avem cite (o ecuatie comuna), apoi le simplificam si inlocuim pe (X).