Por favor preciso de ajuda. (A) 5/2 (B) 1 (A) 2 (B) 1/7 (A) 2/5 (B) 7 (A) 1/7 (B) 5/2 (A) 5/2 (B) 1/.... Pergunta de ideia deRosana2014

Rosana2014 @Rosana2014

July 2022 1 174 Report

Por favor preciso de ajuda.
(A) 5/2

(B) 1

(A) 2

(B) 1/7

(A) 2/5

(B) 7

(A) 1/7

(B) 5/2

(A) 5/2

(B) 1/7

Lista de comentários

lamacch (A) $\lim_{x \to \infty} \dfrac{5 x^{2} +1}{2 x^{2} +3x-7} =$

$\lim_{x \to \infty} \dfrac{ \frac{5 x^{2} }{ x^{2} } + \frac{1}{ x^{2} } }{ \frac{2 x^{2} }{ x^{2} } + \frac{3x}{ x^{2} } - \frac{7}{ x^{2} } } =$

$\lim_{x \to \infty} \dfrac{ 5 + \frac{1}{ x^{2} } }{ 2 + \frac{3}{ x } - \frac{7}{ x^{2} } }=$

$\dfrac{ 5 + \frac{1}{ \infty } }{ 2 + \frac{3}{ \infty } - \frac{7}{ \infty } }=$

$\dfrac{ 5 +0 }{ 2 + 0 - 0 }=$

$\dfrac{ 5 }{ 2 }$

(B) $\lim_{x \to -3} \dfrac{x^{2}+5x +6}{x^{2} -x-12} =$

$\lim_{x \to -3} \dfrac{(x+2).(x+3)}{(x-4).(x+3)} =$

$\lim_{x \to -3} \dfrac{x+2}{x-4} =$

$\dfrac{-3+2}{-3-4} =$

$\dfrac{-1}{-7} =$

$\dfrac{1}{7}$

4 votes Thanks 2

Rosana2014 Obrigaado Lamacch pela ajuda.

lamacch De nada!