can someone explain how to find the limit x--3 using (9-x^2)/(x-3)?

Question

can someone explain how to find the limit x-->3 using (9-x^2)/(x-3)?

anonymous · Answer

Think you have this :
7x = 70
How do you answer ?!:)

anonymous · Answer

x=10

anonymous · Answer

Excelent !
Now when we want to find x in your problem can use this way but do you want to know a better way ?!:)
:D

anonymous · Answer

wait if i solve for x is that going to give me the limit as x approaches 3?

anonymous · Answer

Exacly friend !:)
:D

anonymous · Answer

okay thank you!

amistre64 · Answer

that might be a little misleading ... the limit of a function does not care what the value of the function is

amistre64 · Answer

since at x=3, we get 0/0  the value is undetermined.

the limit is defined to be the value that the function approaches, from the left and right.  regardless of the actual value of the function

amistre64 · Answer

in this case, we can algebra the top to provide an equivalent setup that cancels out the bottom; so that we do not possess a divide by zero result

amistre64 · Answer

what does the top factor out to be?  consider that (x-3) is a factor.

phi · Answer

as amistre said, a good step is factor the top in
$$ \frac{(9-x^2)}{(x-3)} $$

notice the top is a "difference of squares",  a^2 - b^2 which factors into (a+b)(a-b)
if you factor you get
$$ \frac{(3+x)(3-x))}{(x-3)} $$
notice that you can cancel (x-3) from the top and bottom (as long as x is not *exactly* 3)
you get 3+x
now let x -->3
what value are you approaching ?

anonymous · Answer

-6! Sweet thanks so much you guys!!!!

phi · Answer

I hope you are not saying  3+3 = -6
don't you mean +6 ?