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myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1So I think you mean (x^225)/(x5) 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
 6 months ago
Best ResponseYou've already chosen the best response.1Basically is there any factoring you can do?

lornbeach
 6 months ago
Best ResponseYou've already chosen the best response.1Yes that's it and simplify iit

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1yep then plug in 5 to resulting function that is continuous at 5

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1you can share your steps with me if you want

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1not 10/0 nope show me what you did to get that

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1that isn't how you factor x^225

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1x^2a^2=(xa)(x+a) you have x^25^2=(x5)(x+5)

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1right but what do yo notice about the top and bottom (As far as factors go)

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1and you have (x+5)(x5)/(x5)

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1right you basically get rid of the discontinuity so you can plug in x=5
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