Question

Limite???w
lim (x^2-1)/
(x→1-)⁡ |x^2-1|
Limite de x^2 -1 dividido pelo módulo de x^2 -1, quando x tende a 1 pela direita.
expliquem, por favor.

Answer 1

Consideremos a seguinte função na vizinhança à esquerda de 1:

$f(x)=\dfrac{x^2-1}{|x^2-1|}\\\\\\ f(x)=\dfrac{(x+1)\cdot (x-1)}{|(x+1)\cdot (x-1)|}\\\\\\ f(x)=\dfrac{(x+1)\cdot (x-1)}{|x+1|\cdot |x-1|}\\\\\\ f(x)=\dfrac{x+1}{|x+1|}\cdot \dfrac{x-1}{|x-1|}$


Tomando o limite de $f,$ temos

$\underset{x\to 1^-}{\mathrm{\ell im}}~\dfrac{x^2-1}{|x^2-1|}\\\\\\ =\underset{x\to 1^-}{\mathrm{\ell im}}~\dfrac{x+1}{|x+1|}\cdot \dfrac{x-1}{|x-1|}~~~~~~\mathbf{(i)}$


Se x tende a 1 pela esquerda, temos

x < 1    ⇒    x – 1 < 0


E portanto, pela definição de valor absoluto,

|x – 1| = – (x – 1)


Portanto, o limite (i) fica

$=\underset{x\to 1^-}{\mathrm{\ell im}}~\dfrac{x+1}{|x+1|}\cdot \dfrac{x-1}{-(x-1)}\\\\\\ =\underset{x\to 1^-}{\mathrm{\ell im}}~\dfrac{x+1}{|x+1|}\cdot (-1)\\\\\\ =\dfrac{1+1}{|1+1|}\cdot (-1)\\\\\\ =\dfrac{2}{|2|}\cdot (-1)\\\\\\ =\dfrac{2}{2}\cdot (-1)\\\\\\ =1\cdot (-1)\\\\\\ =-1\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\underset{x\to 1^-}{\mathrm{\ell im}}~\dfrac{x^2-1}{|x^2-1|}=-1 \end{array}}$


Bons estudos! :-)