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Dec 9th, 2003, 01:47 AM
#1
Thread Starter
Frenzied Member
can anyone confirm this
http://www.cpcc.edu/academic_learnin...Asymptotes.htm
is this a true easy way to find horizontal asymptotes?
NXSupport - Your one-stop source for computer help
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Dec 9th, 2003, 11:51 AM
#2
Lively Member
Yeah there's nothing wrong with that...
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Dec 9th, 2003, 03:16 PM
#3
transcendental analytic
vertical though, for horizontal it would be lim x->+ and -infinity
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 10th, 2003, 12:31 AM
#4
Thread Starter
Frenzied Member
kedman, if you scroll down it has 3 rules for horizontal
NXSupport - Your one-stop source for computer help
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Dec 10th, 2003, 03:03 AM
#5
transcendental analytic
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 11th, 2003, 12:54 AM
#6
The limit answer kedaman posted was more calculus based (and universal). The 3 rules on the bottom of that page are really pretty much just some algebraic representations of the calc. limit answer for polynomials. In fact, the part where they take the ratio of the highest two powers arises from the others not mattering as x approaches positive and negative infinity, something limits tell you.
The time you enjoy wasting is not wasted time.
Bertrand Russell
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