Prove Limit of f(x)=x4: Proving a=a4

[tanzl]

Homework Statement

Prove that lim (f(x) , x=a) = a⁴ where f(x)=x⁴

Homework 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.

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| < $$\delta$$₁ where $$\delta$$₁ 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 $$\delta$$₁ which is a arbitrary positive number. Then by algebraic manipulation, I can obtain an inequality of |x| smaller than $$\delta$$₁ +a . So when I substitute $$\delta$$₁ +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]

Homework Statement

Prove that lim (f(x) , x=a) = a⁴ where f(x)=x⁴

Homework 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.

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| < $$\delta$$₁ where $$\delta$$₁ 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 $$\delta$$₁ which is a arbitrary positive number. Then by algebraic manipulation, I can obtain an inequality of |x| smaller than $$\delta$$₁ +a . So when I substitute $$\delta$$₁ +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| < $$\delta$$₁ where $$\delta$$₁ 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| < $$\delta$$₁ where $$\delta$$₁ 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?

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