Limit problem. Chapter :Precise definition

[nejnadusho]

Homework Statement

Find the limit

lim x$$\rightarrow$$1 (4+x-3(x^3) = 2

find the values of $$\delta$$ that correspond to $$\epsilon$$=1 and

$$\epsilon$$ = 0.1

Homework Equations

The Attempt at a Solution

As much as I know I should do like this.

|(4 + x - 3(x^3)) - 2| < 1 /////// |x-1| < $$\delta$$

|-3(x^3) + x + 2| < 1 so what so on?


I have no idea.
Thank you in advance.]
Please if you can give me aa short explan with the steps.
Thanks thanks.

 

[sutupidmath]

well this does not look that bad at all: The idea is quite simple indeed. Here is what you need to do

$$|4+x-3x^{3}-2|<\epsilon=1$$ frome here using the properties of the absolute value we have:

$$-1<2+x-3x^{3}<1$$ now let this expression be equal to -1, and 1 ,and solve for x. ALso after that use the fact that

$$|x-1|<\delta=>-\delta<x-1<\delta =>1-\delta<x<1+\delta$$. then use these two expression to come up with the value of delta.

Do the exact same steps for epsilon=0.1

This is almost the whole thing, all i have left out are calculations, simply algebra steps. I think you will be fine from here.

Can you take it from here?

 

[nejnadusho]

Not really.

 

[sutupidmath]

An ideal way would be to manage to solve this eq. $$2+x-3x^{3}=1$$ but since there seem to be no obvious solutions, than doing it is going to be a pain. But i think we can proceede this way also, i am not that sure though:
$$-|3x^{3}|=-|x||3x^{2}|\leq-|x||3x^{2}-1|\leq2-|x||3x^{2}-1|\leq|2-x(3x^{2}-1)|<\epsilon=1$$, so $$|3x^{3}|>1=>-1>3x^{3}>1=>-\frac{1}{3}>x^{3}>\frac{1}{3}$$ the corresponging equation

$$x^{3}=\pm \frac{1}{3}$$ will have three roots, one real, and two complex. so let's take just real ones from here we get

$$-\frac{1}{\sqrt[3]{3}}>x>\frac{1}{\sqrt[3]{3}}$$.

Now using the fact that: $$1-\delta<x<1+\delta$$ we get: $$1-\delta=\frac{1}{\sqrt[3]{3}} \ \ \\ \ and \ \ \\ \ \ 1+\delta=-\frac{1}{\sqrt[3]{3}}$$

Now from here you will be able to find delta, and chose the minimum. That is whicheverone is smaller and positive. But as much as i can see there will be only one such.

I hope this works. I am not sure that what i did is entirely correct, because that polynomial of degree 3 complicated the whole thing. Anyhow i think that this is the idea behind these types of problems.

 

[HallsofIvy]

You know that $\lim_{x\rightarrow 1} 4+ x- x^3= 2$ because you know that polynomials are continuous and putting x= 1 makes $4+ x- x^3$ equal to 2. That in turn tells you that x= 1 makes $4+ x- x^3 -2$ equal to 0. And that tells you that x-1 is a factor of $2+ x- x^3$. Then, by dividing, say, you can determine that $2+ x- x^3= (-1)(x- 1)(3x^2+ 3x+ 2)$. If x is close to 1, we can certainly say that x is between 0 and 2: 0< x< 2 so 0< x²< 4, 0< 3x²< 12, 0< 3x< 6 and, finally, 2< 3x²+ 3x+ 2< 12+ 6+ 2= 20 and so 0< |3x²+ 3x+ 2|< 20. That is, |2+ x- x^3|= |x-1||3x^2+ 3x+ 2|< 20|x-1|.
You can make that less than $\epsilon$ by making |x-1| less than $\epsilon/20$ (or 1 whichever is smaller).

 

[sutupidmath]

You know that $\lim_{x\rightarrow 1} 4+ x- x^3= 2$ because you know that polynomials are continuous and putting x= 1 makes $4+ x- x^3$ equal to 2. That in turn tells you that x= 1 makes $4+ x- x^3 -2$ equal to 0. And that tells you that x-1 is a factor of $2+ x- x^3$. Then, by dividing, say, you can determine that $2+ x- x^3= (-1)(x- 1)(3x^2+ 3x+ 2)$. If x is close to 1, we can certainly say that x is between 0 and 2: 0< x< 2 so 0< x²< 4, 0< 3x²< 12, 0< 3x< 6 and, finally, 2< 3x²+ 3x+ 2< 12+ 6+ 2= 20 and so 0< |3x²+ 3x+ 2|< 20. That is, |2+ x- x^3|= |x-1||3x^2+ 3x+ 2|< 20|x-1|.
You can make that less than $\epsilon$ by making |x-1| less than $\epsilon/20$ (or 1 whichever is smaller).

Well this would be great, only if the OP's polynomial wasn't slightly different from yours here. Your work would be valid if the question at issue actually would be $$\lim_{x\rightarrow 1} 4+ x- x^3= 2$$ while it actually is $$\lim_{x\rightarrow 1} 4+ x- 3x^3= 2$$

$$4+x-3x^{3}$$ It is this 3 in front of the cubic that makes things not look that nice.

 

[HallsofIvy]

Oh blast! My eyes are going.

Actually, everything I said is correct for $4+ x- 3x^3$ because that was what I was working with, even though I wrote $4+ x- x^3$ (which obviously does NOT have limit 2 as x goes to 1).
$4+ x- 3x^3= (-1)(x-1)(3x^2+ 3x+ 1)$ and everything I said applies to that.