Question

Find the limit of the function algebraically.
lim(x -> 5)
x^2-25/x-5

Answer 1

So I think you mean (x^2-25)/(x-5)
right?
This expression does not exist at x=5 because when you plug in 5 you get something/0
Now when you have 0/0 that means the limit can exist

Is there any algebra you can perform to force a function that is continuous at 5?

Answer 2

Basically is there any factoring you can do?

Answer 3

Yes that's it and simplify iit

Answer 4

yep then plug in 5 to resulting function that is continuous at 5

Answer 5

you can share your steps with me if you want

Answer 6

would it be 10/0?

Answer 7

not 10/0 nope
show me what you did to get that

Answer 8

(x+25)(x-25)/x-5

Answer 9

that isn't how you factor x^2-25

Answer 10

x^2-a^2=(x-a)(x+a)

you have
x^2-5^2=(x-5)(x+5)

Answer 11

OH (x+5)(x-5)

Answer 12

right but what do yo notice about the top and bottom (As far as factors go)

Answer 13

and you have (x+5)(x-5)/(x-5)

Answer 14

You can cancel

Answer 15

(x+5)

Answer 16

right
you basically get rid of the discontinuity so you can plug in x=5

Answer 17

so 10

Answer 18

right

Answer 19

Thank you!!

Answer 20

np