infinite question..just wondering what to do..

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k you know for math, when you have to find -oo (negative infinity) what do you do? like normall for just oo I would sub in numbers and see where it was going..but what would do for -ve infinite?
 
bittersweet said:
k you know for math, when you have to find -oo (negative infinity) what do you do? like normall for just oo I would sub in numbers and see where it was going..but what would do for -ve infinite?
I'm sorry, but I don't know what this means.

What do you mean by "having" to "find" infininity? What is "like normal"? Are you working with perpendicular ("normal") something or others? Or are you normalizing something? Where was what "going"? What is "minus vee ee" ("-ve")?

Please reply with specifics. Thank you.

Eliz.
 
Oh sorry k i'll start over. When there's an equation you have to find the limit to. and it says as x is approaching -oo (negative infinite) how would you solve the equation? Normally if it was just lim (x is approaching oo) I would sub in 1,2,3,4,etc into the equation to see a sequence. The sequence may turn out something like 1.2..0.3..0.12.. so that would mean that it would be approaching 0. Sorry I think it would be easier if I was able to find an example of this but I was just wondering how to do this type of question since I saw it on an assignment we had to do. It's not in our books.(Infinite Limits)
 
bittersweet said:
as x is approaching -oo (negative infinite)
For one, this kind of language needs to disappear. You MEAN as it increases without bound. It isn't going toward anything.

To solve such a problem, a limit as x increases (or decreases) without bound, you examine the function and evaluate it's properties. For many functions, it becomes easier as you are more acquainted with more types of functions. For functions you have not previously encountered, you need the skills to understand its properties. Derivatives and ratios can be useful.
 
yup, also a really helpful theorem to know for those limits is:

the limit of 1/x as x->oo is 0
 
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