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