Question

resolva os limites

$\lim_{x \to \ 0} \frac{x^{3} }{2x^{2} -x}$


$\lim_{x \to \ 0} \frac{x^{2} + x }{x^{2} - 3x}$

Answer 1

a -

$\lim_{x \to 0} \frac{x^3}{2.x^2-x}\\\lim_{x \to 0} \frac{0^3}{2.0^2-0}\\ \lim_{x \to 0} \frac{0}{0}$

 Logo, vamos tentar fatorar a expressão:

$\lim_{x \to 0} \frac{x^3}{2.x^2-x}\\ \lim_{x \to 0} \frac{x^3}{(2.x-1).x}\\ \lim_{x \to 0} \frac{x^2}{2.x-1}\\ \lim_{x \to 0} \frac{0^2}{2.0-1}\\ \lim_{x \to 0} \frac{0}{-1}\\ \lim_{x \to 0} = 0$

b -

$\lim_{x \to 0} \frac{x^2+x}{x^2-3.x}\\ \lim_{x \to 0} \frac{0^2+0}{0^2-3.0}\\ \lim_{x \to 0} \frac{0}{0}$

 Então, vamos fatorar a expressão novamente:

$\lim_{x \to 0} \frac{x^2+x}{x^2-3.x}\\ \lim_{x \to 0} \frac{x.(x+1)}{x.(x-3)}\\ \lim_{x \to 0} \frac{x+1}{x-3}\\ \lim_{x \to 0} \frac{0+1}{0-3}\\ \lim_{x \to 0} \frac{1}{-3}\\ \lim_{x \to 0} = -\frac{1}{3}$

Dúvidas só perguntar XD

Answer 2

a)

$\lim_{x \to \ 0} \frac{x^{3} }{2x^{2} -x} \\ \lim_{x \to \ 0} \frac{x.x^{2} }{x(2x - 1)} \\ \lim_{x \to \ 0} \frac{x^{2} }{2x - 1} = \frac{ {0}^{2} }{2.0 - 1} = 0$

b)

$\lim_{x \to \ 0} \frac{x^{2} + x }{x^{2} - 3x} \\ \lim_{x \to \ 0} \frac{x(x + 1)}{x(x - 3)} \\ \lim_{x \to \ 0} \frac{ x + 1}{x - 3} = \frac{0+1}{0 - 3} = -\frac{1}{3}$