Dredge-Master was trying to point out that there are some conditions upon using L^Hopital''s Rule, one of them being that:
f(x) -> 0
g(x) -> 0
as x -> c (c in R*)
but he forgot that L^Hopital''s is also applicable if:
g(x) -> inf
as x -> c
This is the form you were thinking of sjelkjd. However you made a mistake initially in assuming that you can just "replace" the 1/inf * inf with 1/x * x. As people have stated, infinity is NOT a number and we don''t know the rates at which it was reached. We should say 1/inf * inf = 1/g(x) * f(x) where f,g -> inf as x -> c. For Agincourts problem f() = 8 * g().
Incidentally 1/x * x is also (1/x) / (1/x) as well as x/x.
Prove me wrong please
quote:
g(x) -> inf
as x -> c
Only problem is in this case it is still ignored as
g(x)->1/inf, not the other was as the function is 1/x.
so it gets ignored anyway.Thanks for backing me up though.
Complex/imaginary numbers and infinity may be fun, but it sure as hell is still confusing after four years at uni.
Beer - the love catalyst
good ol'' homepage
quote:Original post by jermz
This is the form you were thinking of sjelkjd. However you made a mistake initially in assuming that you can just "replace" the 1/inf * inf with 1/x * x.
Hmm...let''s look at my original post:
quote:
since we can''t use infinity, we can replace it with x and we get
limit x-> inf 1/x*x
Take for example the equation (2x+4)/(3x+1) as x->infinity. If you plug in inf. then (2(inf)+4)/(3(inf)+1) well, 2*inf and 3*inf = inf and inf+4 = inf and so on. So inf/inf, and so = 1
But there''s lots of paradoxes associated with infinity. So now, let''s do it another way:
Divide both sides by x = (2 + (4/x))/(3 + (1/x)) as x->inf, then 4/x, 1/x -> 0 so (2+0)/(3+0) = 2/3, which is not equal to 1.
But there''s lots of paradoxes associated with infinity. So now, let''s do it another way:
Divide both sides by x = (2 + (4/x))/(3 + (1/x)) as x->inf, then 4/x, 1/x -> 0 so (2+0)/(3+0) = 2/3, which is not equal to 1.
quote:Take for example the equation (2x+4)/(3x+1) as x->infinity. If you plug in inf. then (2(inf)+4)/(3(inf)+1) well, 2*inf and 3*inf = inf and inf+4 = inf and so on. So inf/inf, and so = 1
You''ve got to be kidding me. For startes that''s not even an equation. And you can''t just "plug in infinity," and 2*inf doesn''t equal 3*inf nor does it equal inf+4 because those values are undefined. Infinity is not a number. Your limit is wrong.
quote:Divide both sides by x = (2 + (4/x))/(3 + (1/x)) as x->inf, then 4/x, 1/x -> 0 so (2+0)/(3+0) = 2/3, which is not equal to 1.
Both sides? Both sides of what?
quote:Original post by Anonymous Poster
Take for example the equation (2x+4)/(3x+1) as x->infinity. If you plug in inf. then (2(inf)+4)/(3(inf)+1) well, 2*inf and 3*inf = inf and inf+4 = inf and so on. So inf/inf, and so = 1
This DOES NOT equal one. Rather, the limit of (2x+4)/(3x+1) as x->inf is equal to 2/3 (LHopital's)
Ooh, what a surprise! It's the same answer you got from the second part!
edit: added sarcasm
[edited by - sjelkjd on April 14, 2002 12:10:51 AM]
First of all,
x/inf and x/0 is actually undefined
And for the original problem, you all use the wrong equation, the right one is:
lim 1/x*x + 1/x*x + 1/x*x + 1/x*x + 1/x*x + 1/x*x + 1/x*x + 1/x*x
x->inf
or...
lim (8/x)*x = 8
x->inf
problem solved =)
Algorithm for problem solving:
1. Write down the problem
2. Think hard
3. Write down the answer
x/inf and x/0 is actually undefined
And for the original problem, you all use the wrong equation, the right one is:
lim 1/x*x + 1/x*x + 1/x*x + 1/x*x + 1/x*x + 1/x*x + 1/x*x + 1/x*x
x->inf
or...
lim (8/x)*x = 8
x->inf
problem solved =)
Algorithm for problem solving:
1. Write down the problem
2. Think hard
3. Write down the answer
I thought I''d throw in my two cents; I''m probably going to repeat what others have said, but this is what I think:
First of all, the OP wrote that his term division ended with
(1 / inf) ^ inf
1. I don''t know how you came up with that. I would''ve thought
inf * (1 / inf)
2. You can''t use inf in your equations. You have to replace them with limits.
IMO, this is true:
lim [x->inf] ((1 / x) ^ x) = lim [x->inf] (1 / (x ^ x)) = 0
This is why the original expression was wrong.
To expand upon Brother Bob''s post, I would write that this problem really is:
lim [n->inf] (sum ( 8 / 2 ^ n) )<br><br>Since 8 / 2 ^ n is constant, then<br><br> = lim [n->inf] (2 ^ n * ( 8 / 2 ^ n) )<br><br> = 8 no matter what you choose for n<br><br>Furthermore,<br><br><BLOCKQUOTE><SPAN CLASS=smallfont>quote:<hr HEIGHT=1 noshade><br>limit x-> inf 1/x*x<br>which is equal to 1(by L''Hopital''s).<br> <hr height=1 noshade></SPAN></BLOCKQUOTE> <br><br>I don''t know if sjelkjd forgot parentheses, but<br>limit x-> inf 1 / (x*x) = 0<br>limit x-> inf (1 / x) * x = 1<br><br>And you don''t need l''Hospital''s rule. FWIW, I learned that l''Hospital''s rule can only be applied when both the numerator and the denominator yield 0 when the limit is evaluated, or whent they both yield infinity.<br><br>Peace,<br><br>Cédric
First of all, the OP wrote that his term division ended with
(1 / inf) ^ inf
1. I don''t know how you came up with that. I would''ve thought
inf * (1 / inf)
2. You can''t use inf in your equations. You have to replace them with limits.
IMO, this is true:
lim [x->inf] ((1 / x) ^ x) = lim [x->inf] (1 / (x ^ x)) = 0
This is why the original expression was wrong.
To expand upon Brother Bob''s post, I would write that this problem really is:
lim [n->inf] (sum ( 8 / 2 ^ n) )<br><br>Since 8 / 2 ^ n is constant, then<br><br> = lim [n->inf] (2 ^ n * ( 8 / 2 ^ n) )<br><br> = 8 no matter what you choose for n<br><br>Furthermore,<br><br><BLOCKQUOTE><SPAN CLASS=smallfont>quote:<hr HEIGHT=1 noshade><br>limit x-> inf 1/x*x<br>which is equal to 1(by L''Hopital''s).<br> <hr height=1 noshade></SPAN></BLOCKQUOTE> <br><br>I don''t know if sjelkjd forgot parentheses, but<br>limit x-> inf 1 / (x*x) = 0<br>limit x-> inf (1 / x) * x = 1<br><br>And you don''t need l''Hospital''s rule. FWIW, I learned that l''Hospital''s rule can only be applied when both the numerator and the denominator yield 0 when the limit is evaluated, or whent they both yield infinity.<br><br>Peace,<br><br>Cédric
don''t know if this has already been mentioned, but i thought i''d throw it out for y''all.
8 = 8
4 + 4 = 8
2 + 2 + 2 + 2 = 8
can be simplified as
8 * 1 = 8
4 * 2 = 8
2 * 4 = 8
1 * 8 = 8
and then to an equation
(8 / n) * n = 8
and a generic equation
(x / n) * n = x
which, if you simplify using gradeschool algebraic principles, you get
x = x
8 = 8
4 + 4 = 8
2 + 2 + 2 + 2 = 8
can be simplified as
8 * 1 = 8
4 * 2 = 8
2 * 4 = 8
1 * 8 = 8
and then to an equation
(8 / n) * n = 8
and a generic equation
(x / n) * n = x
which, if you simplify using gradeschool algebraic principles, you get
x = x
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