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
$0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} < \frac{1}{{{M}}}\\ 0 < \left| {x - a} \right| < \delta \Rightarrow \left|f(x)\right| > M$

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
$0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} < \varepsilon$

Choose $M=\frac{1}{ε}$ > 0, then
$0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} < \frac{1}{{\left| {f(M)} \right|}}\\ 0 < \left| {x - a} \right| < \delta \Rightarrow \left|f(x)\right| > M$

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 $\displaystyle \frac{1}{f(x)}<\varepsilon= \frac{1}{M} \ \ \dots$

 

[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 $\displaystyle \frac{1}{f(x)}<\varepsilon= \frac{1}{M} \ \ \dots$

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 $|x-a| < \delta$, f(x) > M holds.
But f(x) > M means: $\frac{1}{f(x)}< \frac{1}{M}$. 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 $|x-a| < \delta$, f(x) > M holds.
But f(x) > M means: $\frac{1}{f(x)}< \frac{1}{M}$. 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',
$\frac{1}{f(x)} < \epsilon$ 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
$0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} < \varepsilon$
$0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{ {f(x)}}} < \frac{1}{{{M}}}\\ 0 < \left| {x - a} \right| < \delta \Rightarrow f(x) > M$
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.