lim(as x approaches -9) of (x^2-17x+72)/(x+9)
my first step was to factor the top to. {(x-8)(x-9)}/(x+9)
Usually at this point you should be able to cancel out a factor but I think I am going at this problem wrong. Any ideas?
Thanks
No, there is no factor in the numerator that is the same as the factor in the denominator so there is way to cancel. Notice thaty if you simply put x= -9 in the fraction you get $\displaystyle \frac{(-9)^2- 17(-9)+ 72}{-9+ 9}= \frac{81+ 153+ 72}{0}= \frac{305}{0}$ which does not exist ("is infinity"). The limit does not exist.
It is only when you get "0/0" that you can factor and possibly get a limit.
If you expect this to have a limit, you might want to check that you copied the problem correctly!