Question

determine os limites a seguir:

Answer 1

Resposta:

ver abaixo

Explicação passo-a-passo:

oi vamos lá,

a) $\lim_{x \to 2}(3x-2) = 3(2) - 2 = 4$

b) $\lim_{x \to 4}\frac{x^2-16}{x-4} = \lim_{x \to 4}\frac{(x-4)(x+4)}{x-4} =\lim_{x \to 4}(x+4) = 8$

c) $\lim_{x \to \frac{1}{2} }\frac{4x^2-1}{2x-1} = \lim_{x \to \frac{1}{2} }\frac{(2x-1)(2x+1)}{2x-1} = \lim_{x \to \frac{1}{2} } (2x+1)} = 2$

um abraço

Answer 2

Explicação passo-a-passo:

a)

$\sf lim_{x~\to~2}~3x-2$

$\sf =3\cdot2-2$

$\sf =6-2$

$\sf =\red{4}$

b)

$\sf lim_{x~\to~4}~\dfrac{x^2-16}{x-4}$

$\sf =lim_{x~\to~4}~\dfrac{x^2-4^2}{x-4}$

$\sf =lim_{x~\to~4}~\dfrac{(x+4)\cdot(x-4)}{x-4}$

$\sf =lim_{x~\to~4}~x+4$

$\sf =4+4$

$\sf =\red{8}$

c)

$\sf lim_{x~\to~\frac{1}{2}}~\dfrac{4x^2-1}{2x-1}$

$\sf =lim_{x~\to~\frac{1}{2}}~\dfrac{(2x)^2-1^2}{2x-1}$

$\sf =lim_{x~\to~\frac{1}{2}}~\dfrac{(2x+1)\cdot(2x-1)}{2x-1}$

$\sf =lim_{x~\to~\frac{1}{2}}~2x+1$

$\sf =2\cdot\dfrac{1}{2}+1$

$\sf =\dfrac{2}{2}+1$

$\sf =1+1$

$\sf =\red{2}$