Easy Calculus Problem about Limits
i'm in calc B right now, but i'm a little rusty on limits
can anyone do this problem?
limt of x--> - infinity of -x^2 / x
limit of (negative times x^2 divided by x) as x approaches infinity
February 15, 2011
8 Comments • Newest first
@CrownedSorrow: if a limit goes to -inf/inf, you can use L'hopitals rule to diff the limit (provided denominator =/= 0) and take the limit of that.
also, Lim (-x^2/x) = lim (-x) the x cancel out, as they represent the same value.
the trick is to learn how to cancel them out
even limx->pos.inf. (x^2/(x+9999999999)) = lim(x) as, as x becomes extremely large, the 99999999999 becomes insignificant, and be can just equate (x/x+9999999999999)
= 1, just leaving lim(x).
This is [url=http://courses.skule.ca/exams/bulk/20101/mat195s_2010_exam.pdf]REAL CALCULUS BABY[/url]
[quote=Dacarlover]Not DNE, (-)ve infinity is a valid limit . it just cannot be computed, but it can be manipulated to use it as we please.
ok bye basil D:[/quote]
Well, according to my calc professor, if x is leftover on top of the fraction and goes to -inf/inf, DNE is always true.
[quote=CrownedSorrow]It's negative infinity or DNE.[/quote]
Not DNE, (-)ve infinity is a valid limit . it just cannot be computed, but it can be manipulated to use it as we please.
ok bye basil D:
[quote=Magician27]- x^2/x = - x
as x--> - infinity, it should be -(-infinity), so positive infinity right?[/quote]
It's negative infinity or DNE.
[quote=051035]Negative infinity.
Correct me if I'm wrong.
lim x->-inf (-x^2 / x)
lim x->=inf (-x)
(+)ve infinity.[/quote]
100% right, lim x->- inf (-x) = inf.
the explanation you're looking for is probably the rigorous definition of an infinite limit.
[url=http://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit]I cant type special characters on basil, so you'll just have to figure it out yerself[/url]
- x^2/x = - x
as x--> - infinity, it should be -(-infinity), so positive infinity right?
The answer is Does not Exist.
the problem is approaching negative infinity