For limits of rational expressions, i.e. ""

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  if the numerator is positive and 
   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.

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1.  . Both
    numerator and denominator are 0, so we go to step 3.

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

    1|1 -1 -1 -2
     |   1  0 -1   
      1  0 -1 -3  So 1 is not a zero

    We try 2

    2|1 -1 -1 -2
     |   2  2  2   
      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:



    


You do the other one.  The answer is 

Edwin