# Proving limits

• November 29th 2010, 06:20 AM
chr91
Proving limits
Prove (from the definition) that if f(x) tends to L as x tends to a then 2f(x) tends to 2L as x tends to a

Now we can suppose (For all epsilon>0) (There exists delta >0) (For all x E X)
0<|x-a|<delta implies |f(x) - L| < epsilon.

Now we need to prove that 2f(x) tends to 2L as x tends to a.

I'm struggling a bit with how to do this..
• November 29th 2010, 06:30 AM
emakarov
Quote:

Now we can suppose (For all epsilon>0) (There exists delta >0) (For all x E X) 0<|x-a|<delta implies |f(x) - L| < epsilon.
This means that somebody is able to return a $\delta$ for each $\varepsilon$ given to them. It follows that for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon/2$. Indeed, given an $\varepsilon$, give $\varepsilon/2$ to the $\delta$-providing person that exists according to your statement.

Can you finish?
• November 29th 2010, 08:36 AM
chr91
Quote:

Originally Posted by emakarov
This means that somebody is able to return a $\delta$ for each $\varepsilon$ given to them. It follows that for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon/2$. Indeed, given an $\varepsilon$, give $\varepsilon/2$ to the $\delta$-providing person that exists according to your statement.

Can you finish?

No, I can't. I don't quite understand your explanation ..

I am able to do the limit questions where f(x) , the limit and what x tends to is given. I'm just confused by this particular one..
• November 29th 2010, 09:27 AM
emakarov
One has to show:

for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon/2$. (*)

Since $|f(x) - L| < \varepsilon/2$ is equivalent to $|2f(x) - 2L| < \varepsilon$, (*) means that $\lim\limits_{x\to a}2f(x)=2L$.

How do we show (*)? We are given

for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon$. (**)

The latter statement means that there is a jinny that, given any $\varepsilon$, returns a $\delta$ such that for every x bla, bla, bla. Now, you need to do something similar to that jinny to show (*). Suppose somebody gives you an $\varepsilon$. You halve it and give $\varepsilon/2$ to the jinny. True to its task, the jinny gives you back a $\delta$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon/2$. You take this $\delta$ and return it as your own.
• November 29th 2010, 11:16 AM
chr91
Quote:

Originally Posted by emakarov
One has to show:

for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon/2$. (*)

Since $|f(x) - L| < \varepsilon/2$ is equivalent to $|2f(x) - 2L| < \varepsilon$, (*) means that $\lim\limits_{x\to a}2f(x)=2L$.

How do we show (*)? We are given

for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon$. (**)

The latter statement means that there is a jinny that, given any $\varepsilon$, returns a $\delta$ such that for every x bla, bla, bla. Now, you need to do something similar to that jinny to show (*). Suppose somebody gives you an $\varepsilon$. You halve it and give $\varepsilon/2$ to the jinny. True to its task, the jinny gives you back a $\delta$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon/2$. You take this $\delta$ and return it as your own.

I understand that a little more now, thanks.

It makes sense that I need to show (*) but I still don't get how.

Can I say let delta = epsilon/2 ?

So 0<|x-a| < delta implies |f(x) - l| < epsilon/2

Then that is equaivalent to |2f(x) - 2L| < epsilon ? ? ?
• November 29th 2010, 11:47 AM
HallsofIvy
Quote:

Originally Posted by chr91
I understand that a little more now, thanks.

It makes sense that I need to show (*) but I still don't get how.

Can I say let delta = epsilon/2 ?

No, you can't and no one has suggested such a thing.

Since f(x)->L, given any $\epsilon>0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- L|< \epsilon$.
That's the definition of "limit".

What they are saying is that to prove 2f(x)-> 2L, use the $\delta$ that the previous statement gives for $\epsilon/2$ rather than $\epsilon$.

Quote:

So 0<|x-a| < delta implies |f(x) - l| < epsilon/2

Then that is equaivalent to |2f(x) - 2L| < epsilon ? ? ?
That last part is correct but does NOT imply that you use $\delta/2$. The relationship between $\epsilon$ and $\delta$ is NOT in general "linear"!
• November 29th 2010, 11:49 AM
emakarov
Quote:

Can I say let delta = epsilon/2 ?
No. The relationship between epsilon and delta can be very complex depending on f. Epsilon measures the variation of f(x) whereas delta measures the variation of x.

Quote:

So 0<|x-a| < delta implies |f(x) - l| < epsilon/2
I don't see how this follows.

The difficulty here may be in using the same variable name $\varepsilon$ in two different statements. Let me rewrite this.

We are given

for every $\alpha>0$ there exists a $\beta>0$ such that for all $x\in X$, $0<|x-a|<\beta$ implies $|f(x) - L| < \alpha$. (**)

We need to show

for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon/2$. (*)

We start the proof in the standard way. Fix an arbitrary $\varepsilon$. Instantiate $\alpha$ to $\varepsilon/2$ in (**). Then (**) says that there exists a $\beta$ such that $0<|x-a|<\beta$ implies $|f(x) - L| < \varepsilon/2$. So, take $\delta=\beta$. Then for every x, if $0<|x-a|<\delta$, then $|f(x) - L| < \varepsilon/2$, as required to prove (*).
• November 29th 2010, 12:06 PM
chr91
Quote:

Originally Posted by emakarov
No. The relationship between epsilon and delta can be very complex depending on f. Epsilon measures the variation of f(x) whereas delta measures the variation of x.

I don't see how this follows.

The difficulty here may be in using the same variable name $\varepsilon$ in two different statements. Let me rewrite this.

We are given

for every $\alpha>0$ there exists a $\beta>0$ such that for all $x\in X$, $0<|x-a|<\beta$ implies $|f(x) - L| < \alpha$. (**)

We need to show

for every $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$, $0<|x-a|<\delta$ implies $|f(x) - L| < \varepsilon/2$. (*)

We start the proof in the standard way. Fix an arbitrary $\varepsilon$. Instantiate $\alpha$ to $\varepsilon/2$ in (**). Then (**) says that there exists a $\beta$ such that $0<|x-a|<\beta$ implies $|f(x) - L| < \varepsilon/2$. So, take $\delta=\beta$. Then for every x, if $0<|x-a|<\delta$, then $|f(x) - L| < \varepsilon/2$, as required to prove (*).

Thanks so much for your time and effort to help. I understand it now.

I have a similar question though. It's the same question apart from if f(x) tends to L as x tends to a, then |f(x)| tends to |L| as x tends to a..

Is the format exactly the same but instead of epsilon / 2, we have mod of epsilon?
• November 29th 2010, 12:46 PM
emakarov
Quote:

Is the format exactly the same but instead of epsilon / 2, we have mod of epsilon?
Everywhere we assumed that epsilon > 0, so |epsilon| = epsilon. Here you can instantiate alpha in (**) to just epsilon because once |f(x) - L| < epsilon, | |f(x)| - |L| | <= |f(x) - L| < epsilon.
• November 30th 2010, 04:40 AM
HallsofIvy
Quote:

Originally Posted by chr91
Thanks so much for your time and effort to help. I understand it now.

I have a similar question though. It's the same question apart from if f(x) tends to L as x tends to a, then |f(x)| tends to |L| as x tends to a..

Is the format exactly the same but instead of epsilon / 2, we have mod of epsilon?

epsilon is defined as positive so that would really make no sense.