SOLUTION: Prove lim (2x^2 - x - 1) = 2 as x approaches -1 using precise definition of limits.

 

For any given ε > 0, we must find a δ > 0 such that

|(2x² - x - 1) - 2| < ε whenever |x - (-1)|  = |x + 1| < δ 

|(2x² - x - 1) - 2| < ε iff

|2x² - x - 1 - 2| < ε iff

|2x² - x - 3| < ε iff

|(2x-3)(x+1)| < ε

To find the appropriate δ on the interval (-2,0), we know that

|2x-3| < 7

So on the interval (-2,0), 

|(2x-3)(x+1)| < ε iff 



Thus whenever δ < ε/7, then 

|(2x² - x - 1) - 2| < ε 

thus      [PROVED]

Edwin