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

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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?
Basically is there any factoring you can do?
Yes that's it and simplify iit

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Other answers:

yep then plug in 5 to resulting function that is continuous at 5
you can share your steps with me if you want
would it be 10/0?
not 10/0 nope show me what you did to get that
that isn't how you factor x^2-25
x^2-a^2=(x-a)(x+a) you have x^2-5^2=(x-5)(x+5)
OH (x+5)(x-5)
right but what do yo notice about the top and bottom (As far as factors go)
and you have (x+5)(x-5)/(x-5)
You can cancel
right you basically get rid of the discontinuity so you can plug in x=5
so 10
Thank you!!

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