Proving limit statement using delta-epsilon definition

[drawar]

Homework Statement

For all x\in R , f(x)>0 . Using precise definition of limits and infinite limits, prove that \lim_{x\to a}f(x)=\infty if and only if \lim_{x\to a}\frac{1}{f(x)}=0

Homework Equations

The Attempt at a Solution

I know the precise definition of limits and infinite limits but I cannot see how they can be applied in this case. Also this is biconditional statement so I guess I got to deduce 2 sub-proofs before drawing the conclusion. Ok that's all I've thought of this far, any hints would be greatly appreciated, thanks!

 

[SammyS]

Homework Statement

For all x\in R , f(x)>0 . Using precise definition of limits and infinite limits, prove that \lim_{x\to a}f(x)=\infty if and only if \lim_{x\to a}\frac{1}{f(x)}=0

Homework Equations

The Attempt at a Solution

I know the precise definition of limits and infinite limits but I cannot see how they can be applied in this case. Also this is biconditional statement so I guess I got to deduce 2 sub-proofs before drawing the conclusion. Ok that's all I've thought of this far, any hints would be greatly appreciated, thanks!

If \displaystyle \lim_{x\to\,a}\frac{1}{f(x)}=0 \,, then restate this using δ - ε language.

That should get you started.

 

[drawar]

Ok here's my working for the first part, that is if \lim_{x\to a}\frac{1}{f(x)}=0 then \lim_{x\to a}f(x)=\infty.

Since \lim_{x\to a}\frac{1}{f(x)}=0 exists,
if we let ε > 0, then there exists a δ such that
0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} < \varepsilon

Choose M=\frac{1}{ε} > 0, then
0 &lt; \left| {x - a} \right| &lt; \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} &lt; \frac{1}{{{M}}}\\<br /> 0 &lt; \left| {x - a} \right| &lt; \delta \Rightarrow \left|f(x)\right| &gt; M

How could I get rid of the modulus of f(x) in this case?

 

[SammyS]

Ok here's my working for the first part, that is if \lim_{x\to a}\frac{1}{f(x)}=0 then \lim_{x\to a}f(x)=\infty.

Since \lim_{x\to a}\frac{1}{f(x)}=0 exists,
if we let ε > 0, then there exists a δ such that
0 &lt; \left| {x - a} \right| &lt; \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} &lt; \varepsilon

Choose M=\frac{1}{ε} > 0, then
0 &lt; \left| {x - a} \right| &lt; \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} &lt; \frac{1}{{\left| {f(M)} \right|}}\\<br /> 0 &lt; \left| {x - a} \right| &lt; \delta \Rightarrow \left|f(x)\right| &gt; M

How could I get rid of the modulus of f(x) in this case?

It's almost there.

(I'm still talking about the first part of the proof here.)

What is the definition for \lim_{x\to a}f(x)=\infty\ ? In other words how do we show that \lim_{x\to a}f(x)=\infty\ ?

You need to show that, given any M > 0, there exists a δ such the f(x) > M, whenever 0 < |x-a| < δ .

So, to begin the proof, let M > 0.

Then let ε = 1/M.

Since you are assuming that \displaystyle \lim_{x\to a}\frac{1}{f(x)}=0, you can get your δ which makes \displaystyle \frac{1}{f(x)}&lt;\varepsilon= \frac{1}{M} \ \ \dots

 

[drawar]

It's almost there.

(I'm still talking about the first part of the proof here.)

What is the definition for \lim_{x\to a}f(x)=\infty\ ? In other words how do we show that \lim_{x\to a}f(x)=\infty\ ?

You need to show that, given any M > 0, there exists a δ such the f(x) > M, whenever 0 < |x-a| < δ .

So, to begin the proof, let M > 0.

Then let ε = 1/M.

Since you are assuming that \displaystyle \lim_{x\to a}\frac{1}{f(x)}=0, you can get your δ which makes \displaystyle \frac{1}{f(x)}&lt;\varepsilon= \frac{1}{M} \ \ \dots

When it comes to this kind of problem I always try to manipulate f(x) to the form of |x-a| before choosing a δ that works. But in this generalized case how am I supposed to do that?

 

[Tomer]

Drawar, the technique of manipulating the form of |f(x) - L| and so on is a good one for trying to estimate the \epsilon's and \delta's you need. But eventually you have to understand what your goal is, and what you need to find.

As SammyS said:
You begin by having a general M > 0, for which you need to find a suitable \delta, so that for each x satisfying |x-a| &lt; \delta, f(x) > M holds.
But f(x) > M means: \frac{1}{f(x)}&lt; \frac{1}{M}. Can you find such a \delta that will ensure that? I bet you can -- it's very much implied from your assumption on \frac{1}{f(x)}...
Hope this helps!

 

[drawar]

Drawar, the technique of manipulating the form of |f(x) - L| and so on is a good one for trying to estimate the \epsilon's and \delta's you need. But eventually you have to understand what your goal is, and what you need to find.

As SammyS said:
You begin by having a general M > 0, for which you need to find a suitable \delta, so that for each x satisfying |x-a| &lt; \delta, f(x) > M holds.
But f(x) > M means: \frac{1}{f(x)}&lt; \frac{1}{M}. Can you find such a \delta that will ensure that? I bet you can -- it's very much implied from your assumption on \frac{1}{f(x)}...
Hope this helps!

But how should we choose such a δ? In terms of ε or M? Sorry you guys but I don't really get what you said. I thought after letting ε=1/M then the proof is completed.

 

[SammyS]

When it comes to this kind of problem I always try to manipulate f(x) to the form of |x-a| before choosing a δ that works. But in this generalized case how am I supposed to do that?

Well, this is a different sort of problem.

You're not give a specific function to manipulate in the way you describe.

You assume that \displaystyle \lim_{x\to\,a}\frac{1}{f(x)}=0 is true.

Now to show that \displaystyle \lim_{x\to\,a}f(x)=\infty\ , you need to prove that given any M > 0, (no matter how large) , there exists a δ such that f(x) > M whenever 0<|x-a|<0 .

Use 1/M as your ε. Then using this 1/M as ε with \displaystyle \lim_{x\to\,a}\frac{1}{f(x)}=0 find a delta. Show the this δ works with M for \displaystyle \lim_{x\to\,a}f(x)=\infty\ .

 

[Tomer]

But how should we choose such a δ? In terms of ε or M? Sorry you guys but I don't really get what you said. I thought after letting ε=1/M then the proof is completed.

Well - that's exactly it! You need to promise me you can find such a delta - even without specifying it!
So by choosing, as you and others mentioned, ε = 1/M, you can say, according to your assumption, that there exists a δ (let's call it δ'!), so that for every x satisfying |x-a| < δ',
\frac{1}{f(x)} &lt; \epsilon holds...
This δ' is what you need...

 

[drawar]

So the (<=) proof would look like this, right?

Let M > 0 and let ε=1/M > 0
Since \lim_{x\to a}\frac{1}{f(x)}=0 exists, then there exists a δ such that
0 &lt; \left| {x - a} \right| &lt; \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} &lt; \varepsilon
0 &lt; \left| {x - a} \right| &lt; \delta \Rightarrow \frac{1}{{ {f(x)}}} &lt; \frac{1}{{{M}}}\\<br /> 0 &lt; \left| {x - a} \right| &lt; \delta \Rightarrow f(x) &gt; M
this means \lim_{x\to a}f(x)=\infty

 

[Tomer]

Looks perfect to me ;)
You could sum up by saying: for every M > 0 we have therefore found a δ > 0, so that for every x satisfying |x - a| < δ, f(x) > M holds - which makes it perfect and shows that you know what you're doing - but the important thing is that you understood the logic of it!

 

[drawar]

Looks perfect to me ;)
You could sum up by saying: for every M > 0 we have therefore found a δ > 0, so that for every x satisfying |x - a| < δ, f(x) > M holds - which makes it perfect and shows that you know what you're doing - but the important thing is that you understood the logic of it!

Hell yeah thank you so much!
Ok for the (=>) proof, I think just let M=1/ε and do everything backwards. Am I right?

 

[Tomer]

Yup, sounds about right ;) Make sure you formulate it logically and you have it...

 

[drawar]

Thank you guys for all of your kind help and guidance. :) Finally figured it out now.