Question

calcule o limite abaixo!! porfavor ​

Answer 1

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O limite existe se e somente se os limites laterais existem e são iguais.

$\sf{f(x)}=\begin{cases}\sf{\dfrac{1}{2-3x}~se~x\textless-3}\\\sf{\sqrt[3]{x+2}~se~x\ge-3}\end{cases}$

$\displaystyle\sf{\lim_{x \to -3^{+}}f(x)=\lim_{x \to -3}\sqrt[3]{x+2}=\sqrt[3]{-3+2}=\sqrt[3]{-1}=-1}\\\sf{\lim_{x \to -3^{-}}f(x)=\lim_{x \to -3}\dfrac{1}{2-3x}=\dfrac{1}{11}}$

como

$\displaystyle\sf{\lim_{x \to-3^{+}}f(x)\ne\lim_{x \to -3^{-}}f(x)}\\\displaystyle\sf{\lim_{x \to -3}f(x)\not\exists}$