uni calculus (limits) (1 Viewer)

[...]

umm..any help is appreciated..thx!

Suppoose the function y=f(x) is continuous on the closed interval [0,1] and that 1/2<= f(x) <= 3/4. Show that there exists c E (0,1) such that f(c)=c.

key: E =element symbol

thx

 

[KeypadSDM]

It's continuous, and lies within that gap, thus the line x = y must pass through y = f(x)

:. f(x) = x

:. f(c) = c for some c E (0,1)

I'll draw a diagram in a sec...

 

[KeypadSDM]

Here it is...

 

[martin]

slightly more rigorously,

the Intermediate Value Theorem says
if L is a constant and f is continuous on [a,b], if f(a) < L < f(b) or f(a) > L > f(b) then there is a c in (a,b) such that f(c)=L

now let G(x) = f(x)-x

G(0) = f(0) so 1/2 < G(0) < 3/4
G(1) = f(1) - 1 so -1/2 < G(1) < -1/4

now G(0) > 0 > G(1)

so by IVT there exists c in (0,1) such that G(c)=0

or f(c)-c=0
so f(c)=c

thanks,
Martin

 

[turtle_2468]

btw, you spell it "weird"... If in doubt just check your weird short bow of evisceration.

 

[Archman]

haha someone is playing too much D2

 

[Grey Council]

lol, thats David allright. Too much Diablo 2, not enough maths.

 

[turtle_2468]

yes. But I can't afford one...
and talk to write.. sigh

 

[...]

hehe..heres another one..

using squeezing principle..or pinching ..

lim_{(x-->infinity)} cos(x)/sqrt(x)

hmm..this is my method

-1 <= cos(x) <= 1
-1/sqrt(x) <= limit <= 1/sqrt(x)

now wat happens :S
cos it is sqrt(infinity)

thanks

 

[ND]

1/sqrt(oo)=-1/sqrt(oo)=0

 

[...]

so hmm...

interesting, so sqrt(oo) = 0?

 

[ND]

No, sqrt(oo)=oo (someone correct me if i'm wrong cos the only maths i've done is 4u, i haven't actually done any of this limit stuff in uni), which is why 1/sqrt(oo)=0.

 

[...]

ah opps..big dumb typo there..lol

and yea, i was wondering is sqrt(oo) = oo
cos then 1/sqrt(oo) = 0

 

[KeypadSDM]

Unless you want to get into the hyperreals. Which you don't.

 

[KeypadSDM]

Lim[x->oo] (1/x) = 0

That's the identity you already have, so you can use it.

I.e.

-1 <= Cos[x] <= 1
-1/Sqrt[x] <= Cos[x]/Sqrt[x] <= 1/Sqrt[x]
Lim[x->oo](-1/Sqrt[x]) <= Lim[x->oo](Cos[x]/Sqrt[x]) <= Lim[x->oo](1/Sqrt[x])
0 <= Lim[x->oo](Lim[x->oo]Cos[x]/Sqrt[x]) <= 0
:. Lim[x->oo](Lim[x->oo]Cos[x]/Sqrt[x]) = 0

 

[...]

Originally posted by KeypadSDM
Unless you want to get into the hyperreals. Which you don't.

oh wtf...

 

[...]

Originally posted by KeypadSDM
0 <= Lim[x->oo](Lim[x->oo]Cos[x]/Sqrt[x]) <= 0
:. Lim[x->oo](Lim[x->oo]Cos[x]/Sqrt[x]) = 0

umm..

wtf

2limits??

but even then, does that explain sqrt(x) as x-->oo ???

 

[Affinity]

hmm dunno.. this does though:

to prove as (x-> inf) sqrt(x) -> inf

we only need to show that for any given N>0, there exist M such that whenever x>M, sqrt(x) > N

and M = N^2 fulfills this.

 

[...]

ah..i think i know what u mean affinity..

hmm..
M = N²

ok..u sure its equal???

 

[Affinity]

you can pick any thing not less than N^2 to be M, N^2 is the smallest one which works though