Answer: [tex]\displaystyle \boldsymbol{-\frac{1}{2}}[/tex]
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Work Shown:
[tex]\displaystyle L = \lim_{\text{x}\to 1^{+}} \frac{1-\text{x}}{\text{x}^2-1}\\\\\\\displaystyle L = \lim_{\text{x}\to 1^{+}} \frac{-(\text{x}-1)}{(\text{x}-1)(\text{x}+1)}\\\\\\\displaystyle L = \lim_{\text{x}\to 1^{+}} \frac{-1}{\text{x}+1}\\\\\\\displaystyle L = \frac{-1}{1+1}\\\\\\\displaystyle L = -\frac{1}{2}\\\\\\[/tex]
In the second step, I used the difference of squares rule to factor.
The (x-1) terms cancel which allows us to plug in x = 1. We plug this value in because x is approaching 1 from the right side.
