Question

ALGUÉM BOM EM MATEMATICA PFFFF VALENDO 10 PONTOS, PRECISO DO CALCULO, EU TENHO AS RESPOSTAS

6a) 2/3
b) 5/11
c) 14/5

Answer 1

$a) \lim_{x \to 1} f(x)= \frac{5.(1)^3-6.(1)^2+3.(1)}{(1)^3-(1)^2+3.(1)}= \frac{5.1-6.1+3}{1-1+3}= \frac{5-6+3}{3}= \frac{2}{3}$
$b) \lim_{x \to \frac{1}{2} } f(x)= \frac{5.(\frac{1}{2})^3-6. (\frac{1}{2})^2+3. (\frac{1}{2}) }{ (\frac{1}{2})^3-( \frac{1}{2})^2+3.(\frac{1}{2}) }= \frac{5. \frac{1}{8}-6. \frac{1}{4}+ \frac{3}{2} }{ \frac{1}{8}- \frac{1}{4}+ \frac{3}{2} }= \frac{ \frac{5}{8}- \frac{6}{4}+ \frac{3}{2} }{ \frac{1-2+12}{8} }= \frac{ \frac{5}{8}- \frac{3}{2}+ \frac{3}{2} }{ \frac{11}{8} }$$= \frac{ \frac{5}{8} }{ \frac{11}{8} }= \frac{5}{8}. \frac{8}{11}= \frac{5}{11}$
$c) \lim_{x \to -1} f(x)= \frac{5.(-1)^3-6.(-1)^2+3(-1)}{(-1)^3-(-1)^2+3(-1)}= \frac{5.(-1)-6.1-3}{-1-1-3}= \frac{-5-6-3}{-5}= \frac{-14}{-5}$$= \frac{14}{5}$