x tendendo a 1
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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\\\\\\\](https://tex.z-dn.net/?f=%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%7D++%5Cfrac%7B%28+x%5E%7B2%7D-1%29+%7D%7B%7C%28x-1%29%7C%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%7Cx-1%7C%3D%5Cbegin%7Bcases%7D%0A%28x-1%29%7E%7Ese%7E%7Ex-1+%5Cgeq+0%5C%5C%5C%5C%5C%5C%0A-%28x-1%29%7E%7Ese%7E%7Ex-1%5C+%5Ctextless+%5C+0%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%0A%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%7D+%5Cfrac%7B%28+x%5E%7B2%7D+-1%29%7D%7B%28x-1%29%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%5E%7B%2B%7D+%7D+%5Cfrac%7B%28+x-1%29.%28x%2B1%29%7D%7B%28x-1%29%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%5E%7B%2B%7D+%7D+%28x%2B1%29%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%5E%7B%2B%7D+%7D+%281%2B1%29%3D2%5C%5C%5C%5C%5C%5C%5C%0A%0A%0A++ "\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\\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%5E%7B-%7D+%7D+%5Cfrac%7B%28+x%5E%7B2%7D+-1%29%7D%7B-%28x-1%29%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%5E%7B-%7D+%7D+%5Cfrac%7B%28x-1%29.%28x%2B1%29%7D%7B-%28x-1%29%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%5E%7B-%7D+%7D+%5Cfrac%7B%28x%2B1%29%7D%7B-1%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%5E%7B-%7D+%7D+%5Cfrac%7B%281%2B1%29%7D%7B-1%7D%5C%5C%5C%5C%5C%5C%5C+%0A%5Cdisplaystyle+%5Clim_%7Bx%5Cto+1%5E%7B-%7D+%7D+%5Cfrac%7B%282%29%7D%7B-1%7D%3D-2%5C%5C%5C%5C%5C%5C%5C+%0A%0A "\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!
Solução!
Como os limites são diferentes ,logo concluímos que não existe limite.
Boa noite!
Bons estudos!
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