Question 1163650
<pre>{{{lim["x->1"](x^3) = 1}}} 

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

|x³ - 1| < &epsilon; whenever |x - 1| < &delta; 

|x³ - 1| < &epsilon; iff

|(x - 1)(x² + x + 1)| < &epsilon; iff

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

|x² + x + 1| < 1  (the value of this increasing function when x=0)

So on the interval (0,2), 

|(x - 1)(x² + x + 1)| < &epsilon; iff 

{{{abs(x-1) < epsilon^""/(x^2+x+1) < epsilon/1}}}

Thus whenever &delta; < &epsilon;, then 

|x³ - 1| < &epsilon; 

thus {{{lim["x->-1"](x^3) = 1}}}     [PROVED]

Edwin</pre>