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= Question 1 =

== i ==

<math>\lim_{x\to1}{\frac{x^2-1}{x+1}} = \lim_{x\to1}{\frac{\cancel{(x+1)}(x-1)}\cancel{x+1}} 
</math>

<math>= \lim_{x\to1} {x-1}</math>

<math>= 0</math>

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