Question

calcule os limites! me ajudem. ​

Answer 1

Resposta:

Explicação passo-a-passo:

a)

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

b)

$\lim_{x \to -4} \dfrac{x^{2}+5x+4}{x^{2}+3x-4}=\lim_{x \to -4} \dfrac{(x+1)(x+4)}{(x-1)(x+4)}=\lim_{x \to -4} \dfrac{(x+1)}{(x-1)}=\dfrac{-4+1}{-4-1}=\dfrac{-3}{-5}=\dfrac{3}{5}$

c)

$\lim_{x \to 4} \dfrac{x^{2}-4x}{x^{2}-3x-4}=\lim_{x \to 4} \dfrac{x(x-4)}{(x+1)(x-4)}=\lim_{x \to 4} \dfrac{x}{(x+1)}=\dfrac{4}{4+1}=\dfrac{4}{5}$

d)

$\lim_{t \to -3} \dfrac{t^{2}-9}{2t^{2}+7t+3}=\lim_{t \to -3} \dfrac{(t+3)(t-3)}{(2t+1)(t+3)}=\lim_{t \to -3} \dfrac{(t-3)}{(2t+1)}=\dfrac{-3-3}{2.(-3)+1}=\dfrac{-6}{-6+1}=\dfrac{-6}{-5}=\dfrac{6}{5}$