1. real analysis limits

(1) Let A on R and let f : A --> R be such that f(x) > 0 for all x on A.

Prove that if lim x -->c f(x) exists and is nonzero, for some c on A, then lim x -->c square root(f(x)) = square root (lim x-->c f(x)).

(2) If f, defined on an interval [a,infinity), is a decreasing function and is bounded below then lim x -->infinity f(x) exists.

2. Originally Posted by varkoume.com

(1) Let A on R and let f : A --> R be such that f(x) > 0 for all x on A.

Prove that if lim x -->c f(x) exists and is nonzero, for some c on A, then lim x -->c square root(f(x)) = square root (lim x-->c f(x)).

(2) If f, defined on an interval [a,infinity), is a decreasing function and is bounded below then lim x -->infinity f(x) exists.
We actualy want to prove:

$\lim_{x\to c}{\sqrt f(x)}= \sqrt m$

For that you must prove: given ε>0 there exists a δ>0 such that:

..........if xεA and 0<|x-c|<δ , then $|\sqrt f(x)-\sqrt m|<\epsilon$. The trick of that case is hiding in the last inequality:

........................... $|\sqrt f(x)-\sqrt m|<\epsilon$.......................1

Ιf we now multiply $|\sqrt f(x)-\sqrt m|$ by $|\sqrt f(x)+\sqrt m|$ and at the same time divide by $|\sqrt f(x)+\sqrt m|$ we get:

$\frac{\ |f(x)-m|}{|\sqrt f(x) + \sqrt m |}$.

Now since f(x)>0 ====> $\sqrt f(x)$>0=====> $\sqrt f(x) +\sqrt m > \sqrt m >0$.

Hence $\frac{\ 1}{|\sqrt f(x) +\sqrt m|}$< $\frac{\ 1}{\sqrt m}$ .......................======>

$\frac{\ |f(x)-m|}{|\sqrt f(x) +\sqrt m|}$< $\frac{\ |f(x)-m|}{\sqrt m}$.

Overall now we have :

$|\sqrt f(x) - \sqrt m|< \frac{\ |f(x) - m|}{|\sqrt f(x) + \sqrt m|} < \frac{\ |f(x) - m|}{\sqrt m}$.................................................. 2

This last inequality is very important to the proof that follows:

Note the work we have done so far can be considered as an aside work.

Now for the proof :

Let ε>0 then $\epsilon.\sqrt m >0$.

But , since $\lim_{x\to c}{f(x)} =m$,then for any E>0 we can find a δ>0 such that :

........................if xεA and 0<|x-c|<δ ,then |f(x)-m|<E...............

Since the above holds for any E>0,it will hold for $\epsilon.\sqrt m=E>0$.

And for that E we can find a δ>0, such that:

...........if xεA and 0<|x-c|< δ,then |f(x)-m|< $\epsilon.\sqrt m$.................................................. .3

Now we are nearly done,so let:

...........xεA and 0<|x-c|<δ.............................................

............then in combination of (2) and (3) we get the desired result:

................................ $|\sqrt f(x) - \sqrt m|<\epsilon$.................................................. .....................

Of course if one can write a more detailed proof the more he/she gets to the inside of the proof,and i think this should be the ultimate goal of every student in analysis,otherwise one may run into the possibility of ,what you may remember one year you may have forgotten the next.

3. thanks a lot..this is very helpful..but i still can't solve the 2nd problem ;/

4. Originally Posted by varkoume.com

(1) Let A on R and let f : A --> R be such that f(x) > 0 for all x on A.

Prove that if lim x -->c f(x) exists and is nonzero, for some c on A, then lim x -->c square root(f(x)) = square root (lim x-->c f(x)).

(2) If f, defined on an interval [a,infinity), is a decreasing function and is bounded below then lim x -->infinity f(x) exists.

We want to prove that:

There exists an m such that: $\lim_{ x\to\infty}{f(x)} = m$

Now since f is bounded below and decreasing ,thru the axiom of continuity of real Nos, there exists a real No m such that:

......for all y belonging to the range of f ,yεR(f), $m\leq y$.................................................. ......................................1

......for all ε>0 ,there exists a yεR(f) ,such that $m\leq y< m+\epsilon$.................................................. .......................2

Now let us prove that this m is the limit of f(x) as x goes to infinity.

To prove that,we must prove that:

For all ε>0 there exists a δ>0,such that:

.........if xε[a, $\infty$) and x>δ ,then |f(x)-m|<ε................

So,let ε>0,then according to (2) THERE exists a yεR(f) and $m\leq y< m +\epsilon$.

But since yεR(f) there exists xε[a, $\infty$),and y=f(x).

Here is the trick of the problem.We put for that x=δ,and since x>0 ,δ>0.

(Note we,for simplicity ,consider the case where a>0).

So far we have prove that:

Given ε>0 ,there exists δ>0.It remains to see what happens :

...............if xε[a, $\infty$) and x>δ...........................

But x>δ====> $x\geq\delta$ and since f is decreasing that implies $f(x)\leq f(\delta)$,and according to (2):

...................... $f(x)\leq f(\delta)< m+\epsilon$.................................................. .............................3

Also according to (1) : $m\leq f(x)$,since f(x)εR(f), which implies that m -ε< f(x).............................................. ........................................4

Combining (3) and (4) we get:

.........m-ε<f(x)<m+ε ,which is equivalent to: |f(x) - m|<ε,which is the desired result.

5. thanks