Anyone know how to prove these? (Function & Limits) #3: 1 ÷ (x3 - 2x - 1) = 0...

[harieche]

Anyone know how to prove these? (Function & Limits) #3: 1 ÷ (x3 - 2x - 1) = 0...

Hello everyone,

I have been doing a past paper and there are some questions that I cannot prove. Can anyone help me please?

Here is the question:
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3. Prove by the definition that : 1 ÷ (x³ - 2x - 1) =0 as Limit (x→∞)

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[oleksandrenko]

Definition: Let f be a function defined on some open interval from a to infinity. Then we say the limit of f(x) as x approaches infinity is L, and we write

if for every number

> 0 there is a corresponding number N such that

whenever x > N.

( Let x so big, that x^3-2x-1>0 )

We have 1/(x^3-2x-1) < 1/x^3, so lets N >= 1/epsilon^(1/3)

than if x > N ( N> 1/epsilon^(1/3) ) we have, that |1/(x^3-2x-1) - 0| < 1/x^3 < 1/(1/epsilon^(1/3) )^3= epsilon.

So, for every epsilon there is a corresponding number N>=1/epsilon^(1/3) such that |1/(x^3-2x-1) - 0| < epsilon

 

[harieche]

Few more questions:

1, How can I know 1/(x^3-2x-1) < 1/x^3 ? Do we have to prove by using Triangle inequality?
2, How to get to 1/epsilon^(1/3) ?

I have gone to |x^3-2x-1| > 1/epsilon so far, but I cannot get rid of the x to the power of 3.

 

[stapel]

Few more questions:

1, How can I know 1/(x^3-2x-1) < 1/x^3 ? Do we have to prove by using Triangle inequality?

You're comparing two fractions, each having the same numerator (in this case, "1"). If one fraction's denominator is bigger (that is, if you're dividing the same value by something bigger), what must be true of the result? Is the resulting fraction (or quotient) bigger or smaller?

2, How to get to 1/epsilon^(1/3) ?

You fiddle around with the math, keeping in mind where you need to end up, and figure out what you need as you go. Since the comparison netted you the expression 1/x^3, and you're needing the difference to be less than epsilon, then where does this lead?

I have gone to |x^3-2x-1| > 1/epsilon so far, but I cannot get rid of the x to the power of 3.

Please reply showing your work. Thank you!

 

[harieche]

Here is my work, I am not sure is that correct or not. Please check

1/(x³-2x-1) = 0 as x→∞

If for every epsilon > 0, there exists x₁
| 1/(x³-2x-1) - 0 | < epsilon , wherever x > x₁
| x³-2x-1 | > 1/epsilonwhen x³-2x-1 > 0 , then x³-2x-1 < x³
so |x³|> | x³-2x-1 | > 1/epsilon
let x > 1/epsilon^(1/3)
because x > x_1, so x₁ >=1/epsilon^(1/3) such that | 1/(x³-2x-1) - 0 | < epsilon

 

[oleksandrenko]

Prove, that 1/(x³-2x-1) → 0 as x→∞

If for every epsilon > 0, there exists x₁ such, that |1/(x³-2x-1) - 0| < epsilon , wherever x > x₁

Let x₁ >=1/epsilon^(1/3), than |1/(x³-2x-1) - 0| < 1/|x1^3|< epsilon.

 

[Ishuda]

Hello everyone,

I have been doing a past paper and there are some questions that I cannot prove. Can anyone help me please?

Here is the question:
_______________________________________________________________________________________

3. Prove by the definition that : 1 ÷ (x³ - 2x - 1) =0 as Limit (x→∞)

_____________________________________________________________________________________

You really should only ask a question one time and not several
http://www.freemathhelp.com/forum/t...or-of-the-limit-proving-lim-x→∞-1-(x3-2x-1)-0

 

[Ishuda]

Prove, that 1/(x³-2x-1) → 0 as x→∞

If for every epsilon > 0, there exists x₁ such, that |1/(x³-2x-1) - 0| < epsilon , wherever x > x₁

Let x₁ >=1/epsilon^(1/3), than |1/(x³-2x-1) - 0| < 1/|x1^3|< epsilon.

Let f(x)= (x³-2x-1)
x1 = 2; 1/(x³-2x-1) = 1/3~.333; 1/x³ = 1/8 ~ 0.125 < .333,
not the other way around

See the other post. In essence, you have problems around where f(x)=0 [1/f(x) goes to infinity] so you have to stay away from there.

 

[stapel]

Here is my work, I am not sure is that correct or not. Please check

1/(x³-2x-1) = 0 as x→∞

If for every epsilon > 0, there exists x₁
| 1/(x³-2x-1) - 0 | < epsilon , wherever x > x₁
| x³-2x-1 | > 1/epsilonwhen x³-2x-1 > 0 , then x³-2x-1 < x³
so |x³|> | x³-2x-1 | > 1/epsilon
let x > 1/epsilon^(1/3)
because x > x_1, so x₁ >=1/epsilon^(1/3) such that | 1/(x³-2x-1) - 0 | < epsilon

I'm not entirely sure what you're saying here...? :shock:

Your book should have given you some sort of rubric or pattern for "limits at infinity" exercises, such as the following:

. . . . .\(\displaystyle \mbox{If, for any }\, \epsilon\, >\, 0,\, \mbox{ there exists }\, \delta\, >\, 0\, \mbox{ such that,}\)

. . . . .\(\displaystyle \mbox{if }\, x\, >\, \delta,\, \mbox{ then }\, \big|f(x)\, -\, L\big|\, <\, \epsilon,\, \mbox{ then we can say}\)

. . . . .\(\displaystyle \displaystyle \mbox{that }\, \lim_{x\, \rightarrow\, \infty}\, f(x)\, =\, L\)

In your case, L = 0 and f(x) = 1/(x^3 - 2x - 1). You are needing some delta "d" so that:

. . . . .when x > d, |1/(x^3 - 2x - 1) - 0| = |1/(x^3 - 2x - 1)| < epsilon

So where does this "need" lead us? Let's see:

. . . . .for x > 2, x^3 - 2x -1 is increasing and positive

(Check the graph, if you're not sure.)

So let's deal only with x-values greater than 2. (Since we're "going to infinity", this is fine.) Also, the given function is a cube, so let's look at related cubes, like (x - 1)^3 and (x + 1)^3. The former, we discover (after fiddling about with scratch paper for a bit), is always larger than the given function for x > 2. So:

. . . . .For all x > 2:

. . . . .-3x^2 + 3x < -2x

. . . . .Then, for all x > 2:

. . . . .x^3 - 3x^2 + 3x - 1 < x^2 - 2x - 1

. . . . .But x^3 - 3x^2 + 3x - 1 = (x - 1)^3, so:

. . . . .(x - 1)^3 < x^3 - 2x - 1

. . . . .Then:

. . . . .1/(x^3 - 2x - 1) < 1/(x - 1)^3

. . . . .We want:

. . . . .For x > d, then |1/(x^3 - 2x - 1)| < epsilon

. . . . .Insert the middle part we've manufactured:

. . . . .For x > d, then |1/(x^3 - 2x - 1)| < |1/(x - 1)^3| < epsilon

. . . . .Then, since each of the latter two values is positive, we can say:

. . . . .1/(x - 1)^3 < epsilon

. . . . .1/epsilon < (x - 1)^3

. . . . .cbrt(1/epsilon) < x - 1

. . . . .We want to relate x to d, so if x > d, then everything else follows.

. . . . .So let d = cbrt(1/epsilon) + 1

Note that the above is the "thinking" that goes into the epsilon-delta proof. It is not the proof itself!

The proof would run something like this:

. . .\(\displaystyle \mbox{Pick }\, \epsilon\, >\, 0,\, \mbox{ and let }\, \delta\, =\, \sqrt[3]{\dfrac{1}{\epsilon}\,}\, +\, 1.\)

. . .\(\displaystyle \mbox{Pick }\, x\, >\, \delta. \mbox{ Then:}\)

. . . . .\(\displaystyle \big|f(x)\, -\, 0\big|\, =\, \bigg|\dfrac{1}{x^3\, -\, 2x\, -\, 1}\bigg|\, <\, \bigg|\dfrac{1}{(x\, -\, 1)^3}\bigg|\)

. . .\(\displaystyle \mbox{Because }\, x\, >\, \sqrt[3]{\dfrac{1}{\epsilon}\,}\, +\, 1,\, \mbox{ then:}\)

. . . . .\(\displaystyle x\, -\, 1\, >\, \dfrac{1}{\sqrt[3]{\strut \epsilon\,}}\)

. . . . .\(\displaystyle \dfrac{1}{x\, -\, 1}\, <\, \sqrt[3]{\strut \epsilon\,}\)

. . . . .\(\displaystyle \dfrac{1}{(x\, -\, 1)^3}\, <\, \epsilon\)

. . .\(\displaystyle \mbox{So, for any }\, x\, >\, \delta,\, \mbox{ we have:}\)

. . . . .\(\displaystyle \big|f(x)\, -\, 0\big|\, <\, \bigg|\dfrac{1}{(x\, -\, 1)^3}\bigg|\, <\, \epsilon\)

. . .\(\displaystyle \mbox{Thus:}\)

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \dfrac{1}{x^3\, -\, 2x\, -\, 1}\, =\, 0\)

This uses both delta and epsilon, and the definition of the limit, and clear logic. This, then, would be a proof.