# Limit as x-> 4 of 3-x / x^2-2x-8

1. Jan 1, 2008

### mooneh

lim 3-x / x^2-2x-8
x-> 4-

2. Jan 1, 2008

### Kaan

since as X -> 4-
the numerator will go to -1
and the denominator will go to 0- because x is just a little less than 4

the fraction -1/0- (when 0 is on the left side of the real axis and really close to 0) will give you infinity

3. Jan 1, 2008

### HallsofIvy

Staff Emeritus
Please use parentheses! What you wrote was 3- (x/x^2)- 2x- 8= 3- 1/x- 2x- 8. What you really meant (I think!) was (3- x)/(x^2- 2x- 8). At= x= 4, the numerator is 3- 4= -1 and the denominator is $4^2- 2(4)- 8= 16- 8- 8= 0$. That much tells you the (two-sided) limit does not exist- the two one-sided limits must be positve and negative infinity. If x< 4, say x= 3, then the denominator is $3^2- 2(3)- 8= 9- 6- 8= - 5$. The crucial point is that it is negative. Since a polynomial can only change signs at points where its value is 0 ($x^2- 2x- 8= (x+ 2)(x- 4)$ that can only happen at x= -2 and x= 4), the denominator must negative for -2< x< 4 and x. The numerator is close to -1 for any x close to 4 so the fraction is positive for all -2< x< 4. That tells you the limit is ____________.