Question

Urgente!!! Questões sobre Limite Laterais!!​

Answer 1

Resolução da questão, veja bem.

Questão 3 - A

$\mathbf{\displaystyle\lim_{x\; \to \;2^{-}}~\dfrac{x+2}{x^2-4}}\\ \\ \\ \mathbf{\displaystyle\lim_{x\; \to \;2^{-}}~\dfrac{\diagup\!\!\!\!(x+2)}{\diagup\!\!\!\!\!{(x+2)}(x-2)}}\\ \\ \\ \mathbf{\displaystyle\lim_{x\; \to \;2^{-}}~\dfrac{1}{(x-2)}}\\ \\ \\ =\mathbf{\dfrac{1}{(2-2)}}\\ \\ \\ =\mathbf{\dfrac{1}{0^-}}\\ \\ \\ \large\boxed{\boxed{\boxed{\boxed{\mathbf{\displaystyle\lim_{x\; \to \;2^{-}}~\dfrac{x+2}{x^2-4}=\mathbf{-~\infty}}}}}}$

Questão 3 - D

$\mathbf{\displaystyle\lim_{x\; \to \;2^{+}}~\dfrac{x+2}{x^2-4}}\\ \\ \\ \mathbf{\displaystyle\lim_{x\; \to \;2^{+}}~\dfrac{\diagup\!\!\!\!(x+2)}{\diagup\!\!\!\!\!{(x+2)}(x-2)}}\\ \\ \\ \mathbf{\displaystyle\lim_{x\; \to \;2^{+}}~\dfrac{1}{(x-2)}}\\ \\ \\ =\mathbf{\dfrac{1}{(2-2)}}\\ \\ \\ =\mathbf{\dfrac{1}{0^+}}\\ \\ \\ \large\boxed{\boxed{\boxed{\boxed{\mathbf{\displaystyle\lim_{x\; \to \;2^{+}}~\dfrac{x+2}{x^2-4}=\mathbf{+~\infty}}}}}}$

Espero que te ajude!

Bons estudos!!