showing that the limit exists

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September 23rd 2007, 01:11 PM
dopi

showing that the limit exists

hi i was wondering if anyone could help me with this problem to do with limits. its a real anaylsis subject.

if g, defined on an interval [0,infinity), is decreasing function and is bounded below

then i want to show that lim x-> infinity g(x) exists.

can anyone help? thanks
September 23rd 2007, 02:02 PM
Jhevon

Quote:

Originally Posted by dopi

hi i was wondering if anyone could help me with this problem to do with limits. its a real anaylsis subject.

if g, defined on an interval [0,infinity), is decreasing function and is bounded below

then i want to show that lim x-> infinity g(x) exists.

can anyone help? thanks

do you recall the defintion of a limit? we can say that the function is decreasing and is bounded, which means that there is some g(x) that the value of the function cannot attain, because it is at the boundary. thus we can say that as x is going to infinity, our distance between each g(x) and the boundary is getting arbitrary close. and of course, we can say this more formally (using the $\epsilon- \delta$ definition of a limit) to show that a limit exists and is in fact the boundary
September 23rd 2007, 04:46 PM
Plato

The set $\left\{ {g(x):x \in [0,\infty )} \right\}$ is bounded below. So let $L = \inf \left\{ {g(x):x \in [0,\infty )} \right\}$ .
Now if $\varepsilon > 0$ then $L + \varepsilon$ is not a lower bound. So $\left( {\exists x_0 } \right)\left[ {L \le g\left( {x_0 } \right) < L + \varepsilon } \right]$ .
Because $g$ is decreasing we have $x > x_0 \quad \Rightarrow \quad L \le g(x) \le g\left( {x_0 } \right) < L + \varepsilon$ .

This means that $x > x_0 \quad \Rightarrow \quad \left| {g(x) - L} \right| < \varepsilon$ .
September 25th 2007, 07:39 AM
dopi

can someone pls check this solution

hi i have another question, which i have attempted fully but i was wondering if someone could check it is right.

QUESTION:
Suppose that lim x-> c f(x) = L where L >0, and that lim x-> c g(x) = infinity.

I want to show that lim x->c f(x)g(x) = infinity

MY SOLUTION

Using properties of limits theorem

let f,g be real functions , lim x-> c f(x)g(x) = L x infinity

where L x infinity = infinity

hence lim x-> c f(x)g(x) = inifinity

as required.

can someone please check this thanks
September 25th 2007, 07:55 AM
Jhevon

Quote:

Originally Posted by dopi

hi i have another question, which i have attempted fully but i was wondering if someone could check it is right.

QUESTION:
Suppose that lim x-> c f(x) = L where L >0, and that lim x-> c g(x) = infinity.

I want to show that lim x->c f(x)g(x) = infinity

MY SOLUTION

Using properties of limits theorem

let f,g be real functions , lim x-> c f(x)g(x) = L x infinity

where L x infinity = infinity

hence lim x-> c f(x)g(x) = inifinity

as required.

can someone please check this thanks

there is a limit theorem that says $\lim f(x)\cdot g(x) = \lim f(x) \cdot \lim g(x)$ , i think you should state that somewhere. (something is nagging me about applying this theorem to infinite limits, but it's probably nothin' :D).

so maybe my proof would go along these lines (i will not be formal):

we wish to find $\lim_{x \to c}f(x)g(x)$ .

by theorem (so-and-so, whatever it is in your book), $\lim_{x \to c}f(x)g(x) = \lim_{x \to c}f(x) \lim_{x \to c}g(x)$

$\Rightarrow \lim_{x \to c}f(x)g(x) = L \lim_{x \to c}g(x) = \infty$ .......i'm pretty sure there's a theorem for this last piece as well

QED.

(i would not write something like $= L \cdot \infty$ , that's just awkward)
September 25th 2007, 09:04 AM
dopi

thanks ..ur theory does make more sense..and i dont think it wud have been as easy as i wrote it.
September 25th 2007, 09:20 AM
ThePerfectHacker

Quote:

Originally Posted by dopi

hi i have another question, which i have attempted fully but i was wondering if someone could check it is right.

QUESTION:
Suppose that lim x-> c f(x) = L where L >0, and that lim x-> c g(x) = infinity.

I want to show that lim x->c f(x)g(x) = infinity

So let $I$ be an open interval containing $c$ for which $f,g$ are defined except possibly at $c$ itself.
Let $x_n$ be a sequence in $I - \{ c\}$ converging to $c$ . Then we know that the sequence $f(x_n)$ converges to $L$ and the sequence $g(x_n)$ converges (abusing terminology) to $\infty$ . But this means $\lim \ f(x_n)g(x_n) = \infty$ because $L>0$ by the rule of sequence limits.
September 25th 2007, 09:42 AM
dopi

Quote:

Originally Posted by ThePerfectHacker

So let $I$ be an open interval containing $c$ for which $f,g$ are defined except possibly at $c$ itself.
Let $x_n$ be a sequence in $I - \{ c\}$ converging to $c$ . Then we know that the sequence $f(x_n)$ converges to $L$ and the sequence $g(x_n)$ converges (abusing terminology) to $\infty$ . But this means $\lim \ f(x_n)g(x_n) = \infty$ because $L>0$ by the rule of sequence limits.

but f(x) is a funtion not a sequence
September 25th 2007, 09:47 AM
ThePerfectHacker

Quote:

Originally Posted by dopi

but f(x) is a funtion not a sequence

I know. I am using the following theorem.

If $f(x)$ is a function defined on the open interval $I$ containing $a$ except possibly at $a$ then $\lim_{x\to a}f(x) = L$ if and only if for every sequence in $I - \{ a \}$ converging to $a$ we have $f(x_n)$ is a sequence coverging to $L$ .
September 25th 2007, 09:53 AM
Jhevon

Quote:

Originally Posted by dopi

but f(x) is a funtion not a sequence

i take it that you haven't studied the sequential approach to limits in your class?
September 25th 2007, 09:55 AM
ThePerfectHacker

The sequencial approach the is best approach by far. That is one of the important reasons to developing the theory of sequences before learning lanything about functions. (But you should not avoid $\delta - \epsilon$ because it is useful too at times).
September 25th 2007, 10:01 AM
Jhevon

Quote:

Originally Posted by ThePerfectHacker

The sequencial approach the is best approach by far. That is one of the important reasons to developing the theory of sequences before learning lanything about functions. (But you should not avoid $\delta - \epsilon$ because it is useful too at times).

I agree with you. but what i was getting at is that i never saw the sequential approach until i took advanced calculus. maybe the poster is in that positition as well, and so is not familiar with your approach