lim x-->9 of (x^2-13x+36)/x-9?

Question

I know its not zero, or either infinity, and am now at a complete loss

Answer 1

(x^2-13x-36)/(x-9) simplifies to (x-4)(x-9)/(x-9)

the two (x-9)s cancel, leaving (x-4)
i'm assuming that the whole equation equalled 0, asthey usually do, so:

x-4=0
x=0+4
x=4

Answer 2

limit(when xapproaches9)of
=(x^2-13x+36)/(x-9)
= [(x-4)(x-9)](x-9)
limit(when x approaches 9) of
(x-4)
=9-4
=5. ANS.

Answer 3

Every time you get something divided by 0, you must factorize it. So in this case, we get 0/0. So start factorizing:
(x^2 - 13x + 36)/(x - 9)

= (x^2 - 9x - 4x + 36)/(x - 9)
= [x(x - 9) - 4(x - 9)]/(x - 9)
= [(x - 9)(x - 4)]/(x - 9)
= (x - 4)

Since x -> 9
x - 4 = 5

Hope this helps

Answer 4

the post above is correct about cancelling the 2 (x-9)s...
but after that, it's a simple limit:
lim x-->9 of (x-4)=9-4=5 (not 4)

Answer 5

lim x->9 (x^2-13x+36)/(x-9)
lim x->9 (x^2-4x-9x+36)/(x-9)
lim x->9 [x(x-4)-9(x-4)]/(x-9)
lim x->9 (x-4)(x-9)/(x-9)
lim x->9 x-4
9-4=5(Answer)
is was a good question