Question 1163529
<pre>
{{{lim["x->-1"](2x^2 - x - 1) = 2}}} 

For any given &epsilon; > 0, we must find a &delta; > 0 such that

|(2x² - x - 1) - 2| < &epsilon; whenever |x - (-1)|  = |x + 1| < &delta; 

|(2x² - x - 1) - 2| < &epsilon; iff

|2x² - x - 1 - 2| < &epsilon; iff

|2x² - x - 3| < &epsilon; iff

|(2x-3)(x+1)| < &epsilon;

To find the appropriate &delta; on the interval (-2,0), we know that

|2x-3| < 7

So on the interval (-2,0), 

|(2x-3)(x+1)| < &epsilon; iff 

{{{abs(x+1) < epsilon/((2x-3)) < epsilon/7}}}

Thus whenever &delta; < &epsilon;/7, then 

|(2x² - x - 1) - 2| < &epsilon; 

thus {{{lim["x->-1"](2x^2 - x - 1) = 2}}}     [PROVED]

Edwin</pre>