Lim x--> + or - infinity f(x)

[Tim_Lou]

assume that a funtion converges,

 

that means the limit of the function as it approaches infinity exists.....

 

but lets say,

as x approaches infintity from the left, its gonna be the same,

what about from the right? the function doesnt even exist beyond infinity...!??!?

 

so, it doesnt agree and no limit exist as x approaches infinity? (or - infinity)

 

???????

im confused.

 

 

if it exists, what about function like 1/sqr(1-x^2) or some function whos domain is limited to a certain real number....

 

look, whats the lim of x--> 1 [1/sqr(1-x^2)], infinity?

but there is no graph on the right? the limit approaching from the right is undefined, does the limit exist or not???

[alexander]

1/sqr(1-x^2)...

but there is no graph on the right? the limit approaching from the right is undefined, does the limit exist or not???

in short, there is no limit of this function, let me explain why...

in this function 1 is devided by a valie that is influenced by x, from math we know that the smaller the number that 1 is devided by, the larger value you get ex:

1/1 = 1

1/.1=10

1/.0001 = 10000

in other words, as the denominator approaches zero, you get closer and closer to the infinity.

So, lets jump in and find out what will happen with y as we change x in your function:

x=-4, y=.45

x=-3, y=.5

x=-2, y=.58

x=-1, y=.7

x=0, y=1

now so far there hasnt been anything crazy, but as x approaches 1, y will skyrocket

x=.5, y=1.4

x=.9999001, y=100.5

x=.9999901, y=317.82

x=.9999991, y=1054.1

if x=1, y=infinity because 1/0=infinity

[Tim_Lou]

so, infinity isnt a limit?

 

............

 

oh yeah, a limit of infinity would be pretty contracdicting.

 

well, let me restat my question then:

what is the limit of this function?

f(x)={f(x)=1 x>0}

 

does lim x--> 0 exist?

lim x-->0+ = 1

but lim x-->0-=nothing!

(there is no graph on the left!)

 

lim x--> 0+ doesnt equal lim x--> 0-

which is against the whatever theorm.......

[Bo]

as x approaches infintity from the left, its gonna be the same,

what about from the right? the function doesnt even exist beyond infinity...!??!?

i am quite sure it is possible to define the limit from above to infinity. Basicly because we can take infinity as large as we want to. So defining something larger then infinity as a sensible reference point is useless; because without problems we could shift infinity above that point.

 

look, whats the lim of x--> 1 [1/sqr(1-x^2)], infinity?

but there is no graph on the right? the limit approaching from the right is undefined, does the limit exist or not???

 

No it doesn't; the sqrt of a negative number is a complex number. since you assumed only real numbers, we cant take a limit from above to 1.

(btw what i think alexander means is that this function doesn;t converge for x->1; you can always take a limit)

 

so, infinity isnt a limit?

No, infinity is a completely valid limit. A simple example (i use / for the limit from below and / for the limit from above): (lim x/0) 1/x = -inf. and (lim x/0) 1/x = +inf; while (lim x->+inf) 1/x = (lim x->-inf) 1/x = 0. (Tormod: when will we get an equation editor on these fora? )

 

lim x--> 0+ doesnt equal lim x--> 0-

which is against the whatever theorm.......

Well there is no theorem that says that. There are definitions that use this criterium. e.g. A so called 'continuous function' Should have for every point A that lim x/a = lim x/a (in sloppy notation ) This basicly means that the function is 'nice and smooth' (no strange jumps or gaps)

 

Bo

[Tim_Lou]

so, there is not limit at the end points of a function, right?

[Bo]

so, there is not limit at the end points of a function, right?

well not quite true; if we have a function defined on the interval [0,1] then we can take the limit from above to 0 and from below to 1, mut we can't take the limit from below to 0 (or from above to 1); because the function isn't defined there.

 

Bo

[hashamgondal]

when u write

Lim

x--> infinity

 

then u write a very false statement

if x is aproaching to infinity then the limit of the x is not possible because infinity is limitless

as this is not true to write the limit with infinity

so the above mentioned statement is wrong

[sanctus]

Welcome to the forum.

I don't agree with what you are saying. The infinity is a concept not a number, you are right saying that one can't reach infinity.But one can study what happens when you aproach (not reach!) infinity. Therefore you can calculate that limit.

 

Anyway either I didn't understand what you said or you think that you found a paradox in something that millions of peoples uses/has used in about 1.5 centuries (since Cauchy).

 

In the first case explain what you mean

Inthe second case I would suggest you look a bit more into the theory before putting an angry smiley in the title.

[C1ay]

when u write

Lim

x--> infinity

 

then u write a very false statement

if x is aproaching to infinity then the limit of the x is not possible because infinity is limitless

as this is not true to write the limit with infinity

so the above mentioned statement is wrong

 

Just because the limit cannot be evaluated at infinity does not mean that the limit does not exist. Continuity for example is not a requirement for a limit and that is what you are implying.

[maddog]

when u write

Lim

x--> infinity

...

if x is aproaching to infinity then the limit of the x is not possible because infinity is limitless

as this is not true to write the limit with infinity

so the above mentioned statement is wrong

The definition for "x --> infinity" is "as x is approaching infinity" [forgive me that I don't use the symbol]

 

The limit exist ONLY when being evaluate at ever closer values to the limiting value. If the difference

of each successive approximation is less and less (Cauchy sequence) then there is convergence.

Otherwise there is divergence. Look up in any Calculus book. There will be a section Series & Sequences

& Limits.

 

The example I was thinking of is where f(x) = tan(x)

 

Lim tan(x) = infinity

x --> pi/2+

 

Lim tan(x) = - infinity

x --> pi/2-

 

So here both limits diverge and do not approximated each other.

 

Lim tan(x) = infinity == ??

x --> infinity

 

This limit is meaningless {it can be (- infinity, + infinity) }

 

Maddog

[hashamgondal]

Just because the limit cannot be evaluated at infinity does not mean that the limit does not exist. Continuity for example is not a requirement for a limit and that is what you are implying.

 

i'm not saying that limit does not exist

limit exists but not with infinity

if u r going to draw real numbers on graph then u will start from zero to +infinity proceeding right side and zero to -infinity proceeding left side

so here u dont know where from numbers are comming to center zero and where to go after zero

by this there would be no limit for real numbers approaching to infinity from both sides in graph

[maddog]

there would be no limit for real numbers approaching to infinity from both sides in graph

Actually, this is not true.

The real number line (Reals) from - infinity to + infinity can be mapped ONTO the Complex Unit Circle

where "infinity" at both ends join. These are isomorphically the same. So you approach the limit from

either end. Only with the mapping to the Unit Circle do you get the end points to equal.

 

In Complex Z^2 = x^2 + (iy)^2 = 1

Well in XY plane this is the same as

 

sin^2(x) + cos^2(y) = 1 in Real XY plane.

 

It is true in you case {- infinity, + infinity} is an open interval while [0, 2*pi) is truely a closed interval

(no duplicates). Enjoy! :cup:

 

Maddog