Question 834962
<pre>

{{{matrix(2,2,

  lim, ((x^3-x^2-x-2)/(x^2-6x+8)),  
"x->2","")}}}

For limits of rational expressions, i.e. "{{{(a_polynomial)/(a_polynomial)}}}"

1. Substitute what x approaches.  If the denominator is not 0. 
   what you get is the answer, and you're done.

2. If the denominator is 0, and the numerator is not 0, then the    
   answer is {{{infinity}}} if the numerator is positive and {{{-infinity}}}
   if the numerator is negative, and you're done.

3.  If the numerator is also 0, factor numerator and denominator and
    cancel all common factors. 

4. Substitute what x approaches, simplify, and what you get is the answer.

------------------------

{{{matrix(2,2,

  lim, ((x^3-x^2-x-2)/(x^2-6x+8)),  
"x->2","")}}}

1.  {{{(2^3-2^2-2-2)/(2^2-6(2)+8)=(8-4-2-2)/(4-12+8)=0/0}}}. Both
    numerator and denominator are 0, so we go to step 3.

3.  Factor the numerator: Candidates for zeros are ±1,±2
    {{{x^3-x^2-x-2}}} has 1 sign change, so it has one 
    positive zero.  We try 1:

    1|1 -1 -1 -2
     |<u>   1  0 -1</u>   
      1  0 -1 -3  So 1 is not a zero

    We try 2

    2|1 -1 -1 -2
     |<u>   2  2  2</u>   
      1  1  1  0  So 2 is a zero, so the numerator factors
                  as (x-2)(x²+x+1)

    The denominator x²-6x+8 factors as (x-2)(x-4)

So we have:
{{{matrix(2,2,
  lim, ((x^3-x^2-x-2)/(x^2-6x+8)),  
"x->2","")}}}{{{""=""}}}{{{matrix(2,2,  lim, (((x-2)(x^2+x+1))/((x-2)(x-4))),  
"x->2","")}}}{{{""=""}}}

{{{matrix(2,2,  lim, (((cross(x-2))(x^2+x+1))/((cross(x-2))(x-4))),  
"x->2","")}}}{{{""=""}}}{{{matrix(2,2,  lim, ((x^2+x+1)/(x-4)),  
"x->2","")}}}{{{""=""}}}
    
{{{(2^2+2+1)/(2-4)}}}{{{""=""}}}{{{(4+2+1)/(-2)}}}{{{""=""}}}{{{-7/2}}}

You do the other one.  The answer is {{{16/3}}}

Edwin</pre>