Definition of Limit

[tanzl]

1. The problem statement, all variables and given/known data
Prove that lim (f(x) , x=a) = a4 where f(x)=x4


2. Relevant equations
I think I understand the mechanics of limit proof. I just want to improve my way of presenting it because sometimes even I feel ambiguous for my own works.


3. The attempt at a solution
For all \epsilon>0, there exists \delta>0 such that if 0 < |x-a| < \delta, then |f(x)-a4| < \epsilon

I think the mechanic of the proof is I need to find delta in terms of epsilon such that it satisfies all the conditions above. (or another way round, for me, the proof more like I need to express f(x)-L where L is the limit in terms of delta so that I could make it smaller than epsilon... please correct me)

So, the first thing I need to do is to express |f(x)-a4| in terms of delta.
|f(x)-a4|
=|x^4 - a^4|
=|x-a| * |x3 + ax2 + a2x + a3|
<=|x-a| * |x3| + |ax2| + |a2x| + |a3|

My problem comes, (when it starts to sound ambiguous)
What I do is I will let |x-a| < \delta1 where \delta1 is a positive number such that |x3| + |ax2| + |a2x| + |a3| < P where P is a positive number.

My reason of doing this is I need to obtain a close interval for the expression |x3| + |ax2| + |a2x| + |a3| so that I can make |f(x) - L| smaller than epsilon. Therefore, I started out by letting |x-a| smaller than \delta1 which is a arbitrary positive number. Then by algebraic manipulation, I can obtain an inequality of |x| smaller than \delta1 +a . So when I substitute \delta1 +a into each terms of |x3| + |ax2| + |a2x| + |a3|, I can get inequality of each terms smaller than something. To save the trouble, I let the whole expression less than P which is also an arbitrary positive number.

Since |x3| + |ax2| + |a2x| + |a3| < P
then |x-a| * |x3| + |ax2| + |a2x| + |a3| < \delta *P

So, when I let \delta *P = \epsilon. (I should express delta in terms of epsilon or epsilon in terms of delta?? The first sounds more correct for me because our mission is to find a delta for any given epsilon)

|f(x)-a4| < \epsilon

Done.

Please feel free to correct me.

[konthelion]

I'll rewrite what you've wrote thus far.

\forall \epsilon > 0, \exists \delta > 0 such that \left| x^4 - a^4 \right| < \epsilon whenever 0 < \left| x -a \right| < \delta

From here, we need:

\left| x^4 - a^4 \right| < \epsilon

Factor we get:

\left| x - a \right| \left| x^3 + a^{2}x+x^{2}a+a^3 \right| < \epsilon

Since we have that \delta > \left| x - 2 \right| , then you need to find a delta such that

\left| x - a \right| \left| x^3 + a^{2}x+x^{2}a+a^3 \right| < \left| x - a\right|*P < \delta*P < \epsilon

Naturally, you want such that \delta = \frac{\epsilon}{P}

Hope that helps.

[HallsofIvy]

1. The problem statement, all variables and given/known data
Prove that lim (f(x) , x=a) = a4 where f(x)=x4


2. Relevant equations
I think I understand the mechanics of limit proof. I just want to improve my way of presenting it because sometimes even I feel ambiguous for my own works.


3. The attempt at a solution
For all \epsilon>0, there exists \delta>0 such that if 0 < |x-a| < \delta, then |f(x)-a4| < \epsilon

I think the mechanic of the proof is I need to find delta in terms of epsilon such that it satisfies all the conditions above. (or another way round, for me, the proof more like I need to express f(x)-L where L is the limit in terms of delta so that I could make it smaller than epsilon... please correct me)

So, the first thing I need to do is to express |f(x)-a4| in terms of delta.
|f(x)-a4|
=|x^4 - a^4|
=|x-a| * |x3 + ax2 + a2x + a3|
<=|x-a| * |x3| + |ax2| + |a2x| + |a3|
You need parentheses, of course. But there is no reason to break up |x3+ ax2+ a2+ a3|. It's awkward and doesn't help.

My problem comes, (when it starts to sound ambiguous)
What I do is I will let |x-a| < \delta1 where \delta1 is a positive number such that |x3| + |ax2| + |a2x| + |a3| < P where P is a positive number.

My reason of doing this is I need to obtain a close interval for the expression |x3| + |ax2| + |a2x| + |a3| so that I can make |f(x) - L| smaller than epsilon. Therefore, I started out by letting |x-a| smaller than \delta1 which is a arbitrary positive number. Then by algebraic manipulation, I can obtain an inequality of |x| smaller than \delta1 +a . So when I substitute \delta1 +a into each terms of |x3| + |ax2| + |a2x| + |a3|, I can get inequality of each terms smaller than something. To save the trouble, I let the whole expression less than P which is also an arbitrary positive number.

Since |x3| + |ax2| + |a2x| + |a3| < P
then |x-a| * |x3| + |ax2| + |a2x| + |a3| < \delta *P

So, when I let \delta *P = \epsilon. (I should express delta in terms of epsilon or epsilon in terms of delta?? The first sounds more correct for me because our mission is to find a delta for any given epsilon)

|f(x)-a4| < \epsilon

Done.

Please feel free to correct me.
You are assuming, in any case, that x is close to a, say a-1< x< a+1. x2< (a+1)2 and x3< (a+ 1)3. That should let you set an upperbound on |x3+ ax2+ a2+ a3|.

[tanzl]

It helps a lot. Thank you.

I have a paradox here. Please tell me what is wrong.
I need to prove that limit (f(x) , x = a-) = -\infty
f(x) = \frac{x}{(x-1)^2(x-3)}

1st case

For all M , where M is a arbitrary large number, there exists \delta>0 such that f(x)<M whenever 0 < a-x < \delta

0 < a-x < \delta.
a-\delta < x < a
Therefore x<a ....(1)

Let |x-a| < \delta1 where \delta1 is a positive number such that \frac{1}{(x-1)^2(x-3)} < P where P is a positive number. ...(2)

f(x) = \frac{x}{(x-1)^2(x-3)}
< \frac{a}{P} .... from (1) and (2)
Let \frac{a}{P}=M
f(x) < M

Done.

2nd case

For all M , where M is a arbitrary large number, there exists \delta>0 such that f(x)>M whenever 0 < a-x < \delta
I proved this for fun at first. But, it bothers me when it became confusing that the proof contradicts itself.

As above,
0 < a-x < \delta.
a-\delta < x < a
Therefore x>a-\delta ....(1)

Let |x-a| < \delta1 where \delta1 is a positive number such that \frac{1}{(x-1)^2(x-3)} > P where P is a positive number. ...(2)

f(x) = \frac{x}{(x-1)^2(x-3)}
> \frac{a-\delta}{P} .... from (1) and (2)
Let \frac{a-\delta}{P}=M
f(x) > M

Done.

I derive the both cases with the same arguments but obtain opposite result. What is wrong with my proof and how valid this proof can be?

How to type limit notation by LATEX...