# Math Help - showing that the limit exists

1. ## 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

2. 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

3. 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 $x > x_0 \quad \Rightarrow \quad \left| {g(x) - L} \right| < \varepsilon$.

4. ## 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

5. 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' ).

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)

6. thanks ..ur theory does make more sense..and i dont think it wud have been as easy as i wrote it.

7. 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.

8. 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

9. 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 $L$.

10. 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?

11. 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).

12. 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