## lornbeach one year ago Find the limit of the function algebraically. lim(x -> 5) x^2-25/x-5

1. myininaya

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?

2. myininaya

Basically is there any factoring you can do?

3. lornbeach

Yes that's it and simplify iit

4. myininaya

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

5. myininaya

you can share your steps with me if you want

6. lornbeach

would it be 10/0?

7. myininaya

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

8. lornbeach

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

9. myininaya

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

10. myininaya

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

11. lornbeach

OH (x+5)(x-5)

12. myininaya

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

13. myininaya

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

14. lornbeach

You can cancel

15. lornbeach

(x+5)

16. myininaya

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

17. lornbeach

so 10

18. myininaya

right

19. lornbeach

Thank you!!

20. myininaya

np