Question

x tendendo a 1$\lim_{x \to \1} \frac{ x^{2} -1}{|x-1|}$

Answer 1

Boa noite Evanildo!

Solução!

 

$\displaystyle \lim_{x\to 1} \frac{( x^{2}-1) }{|(x-1)|}\\\\\\\\\\ |x-1|=\begin{cases} (x-1)~~se~~x-1 \geq 0\\\\\\ -(x-1)~~se~~x-1\ \textless \ 0 \end{cases}\\\\\\ \displaystyle \lim_{x\to 1} \frac{( x^{2} -1)}{(x-1)}\\\\\\\ \displaystyle \lim_{x\to 1^{+} } \frac{( x-1).(x+1)}{(x-1)}\\\\\\\ \displaystyle \lim_{x\to 1^{+} } (x+1)\\\\\\\ \displaystyle \lim_{x\to 1^{+} } (1+1)=2$



$\displaystyle \lim_{x\to 1^{-} } \frac{( x^{2} -1)}{-(x-1)}\\\\\\\ \displaystyle \lim_{x\to 1^{-} } \frac{(x-1).(x+1)}{-(x-1)}\\\\\\\ \displaystyle \lim_{x\to 1^{-} } \frac{(x+1)}{-1}\\\\\\\ \displaystyle \lim_{x\to 1^{-} } \frac{(1+1)}{-1}\\\\\\\ \displaystyle \lim_{x\to 1^{-} } \frac{(2)}{-1}=-2$

Como os limites são diferentes ,logo concluímos que não existe limite.

Boa noite!
Bons estudos!