Limits - Prove that xsin1/x approaches 0 near 0

[Miike012]

limits -- Prove that xsin1/x approaches 0 near 0

Prove that xsin1/x approaches 0 near 0.

similar Proof from book
|sin1/x| ≤ 1
| xsin1/x | ≤ |x| for all x not equal to 0, so we can make |xsin1/x|< ε by requiring that |x| < ε and not equal to 0.

MY QUESTION: Prove x²sin1/x approaches 0 near 0.

According to the book, if ε>0, to ensure that |x²sin1/x |< ε we need only require that |x| < ε and not equal to 0.

shouldnt |x| be less that √ε? Hence |x² | < ε and |x|< √ε.

 

[DryRun]

Use the Pinching theorem.
$$-x \le x\sin \frac{1}{x} \le x$$

 

[Miike012]

Use the Pinching theorem.
$$-x \le x\sin \frac{1}{x} \le x$$

Ok how does that tell me that |x²sin1/x|< ε for |x|< ε??

 

[SammyS]

Prove that xsin1/x approaches 0 near 0.

similar Proof from book
|sin1/x| ≤ 1
| xsin1/x | ≤ |x| for all x not equal to 0, so we can make |xsin1/x|< ε by requiring that |x| < ε and not equal to 0.

MY QUESTION: Prove x²sin1/x approaches 0 near 0.

According to the book, if ε>0, to ensure that |x²sin1/x |< ε we need only require that |x| < ε and not equal to 0.

shouldnt |x| be less that √ε? Hence |x² | < ε and |x|< √ε.

Are you supposed to use a δ - ε proof, or can you use such things as the "squeeze theorem" ?

 

[Miike012]

Are you supposed to use a δ - ε proof, or can you use such things as the "squeeze theorem" ?

delta epsilon proof

 

[Miike012]

Prove that x^2sin1/x approaches 0 near 0.


According to the book, if ε>0, to ensure that |x²sin1/x |< ε we need only require that |x| < ε and not equal to 0.

shouldnt |x| be less that √ε? Hence |x² | < ε and |x|< √ε.

all I am doing is quoting from the book... and I want to know how they decided that |x|<ε

 

[Boorglar]

Well, for small enough $\epsilon$, $0<\epsilon < \sqrt{\epsilon}$.

In fact, we only need $0<\epsilon<1$ for this to be true.

Oh and also for a more fundamental reason. In Spivak's book, (I don't know for other books) the proof that this limit is 0 using delta-epsilon comes before the proof that every positive number has a square root (which requires the least upper bound property). So the existence of such a number $\sqrt{\epsilon}$. cannot be assumed.

 

[gopher_p]

This is a common hangup for people doing $\epsilon-\delta$ proofs for the first time.

You don't need to find the "best" $\delta$; you just need to find one that works.

 

[SammyS]

If $\displaystyle |x| < 1\,,\ \text{ then (multiplying by }|x| \text{ gives } |x^2|<|x|\,,\text{ and } |x| < 1\, \text{ ) } |x^2|<1 \ .$

So let δ = min(1,√ε) .

 

[Dickfore]

There is a general theorem:
If $f(x) \rightarrow 0, \ x \rightarrow a$, and $\vert g(x) \vert \le M, \ \forall x \in U(a)$, where $U(a)$ is some neighborhood of the point $a$, then:
$$\mathrm{\lim}_{x \rightarrow a} f(x) \, g(x) = 0$$

Can you prove it?

 

[e^(i Pi)+1=0]

I just learned the δε definition myself and am very shaky on it. Since |x|< δ can we not turn |x²sin($\frac{1}{x}$)|< ε into |x|<$\frac{ε}{xsin(\frac{1}{x})}$. And now we have our δ=$\frac{ε}{xsin(\frac{1}{x})}$ and use this in the proof.

all I am doing is quoting from the book... and I want to know how they decided that |x|<ε

Because it's part of the εδ definition that |x²sin(1/x)|<ε and it's just the nature of the function that |x| ≤ |xsin(1/x)|

 

[micromass]

I just learned the δε definition myself and am very shaky on it. Since |x|< δ can we not turn |x²sin($\frac{1}{x}$)|< ε into |x|<$\frac{ε}{xsin(\frac{1}{x})}$. And now we have our δ=$\frac{ε}{xsin(\frac{1}{x})}$ and use this in the proof.

No, this way our δ would depend on x.

The ε-δ definition tells us that "For all ε, there must exist a δ, such that for all x, ...". This means that our δ must work for all x. Your proposed δ does not work for all x.

 

[HallsofIvy]

For $0< \epsilon< 1$, $\epsilon< \sqrt{\epsilon}$ so "$<\epsilon$" is sufficient.

 

[gopher_p]

No, this way our δ would depend on x.

The ε-δ definition tells us that "For all ε, there must exist a δ, such that for all x, ...". This means that our δ must work for all x. Your proposed δ does not work for all x.

You've got your quantifiers mixed up an have given the definition of uniform continuity.

 

[micromass]

You've got your quantifiers mixed up an have given the definition of uniform continuity.

No, I didn't...

Uniform continuity would be "Forall ε>0, there is a δ>0, such that for all x,y..."
Here, I'm holding 0 fixed and I say " "Forall ε>0, there is a δ>0, such that for all x..."

OK, it might be confusing though that I did not write the definition fully.

 

[SammyS]

If $\displaystyle |x| < 1\,,\ \text{ then (multiplying by }|x| \text{ gives } |x^2|<|x|\,,\text{ and } |x| < 1\, \text{ ) } |x^2|<1 \ .$

So let δ = min(1,√ε) .

I must have had a brain cramp!

It looks like you could use either of the following for δ :

δ = √ε

δ = min(1, ε)​

 

[gopher_p]

No, I didn't...

Uniform continuity would be "Forall ε>0, there is a δ>0, such that for all x,y..."
Here, I'm holding 0 fixed and I say " "Forall ε>0, there is a δ>0, such that for all x..."

OK, it might be confusing though that I did not write the definition fully.

Yes. You are correct. I misread your yada-yada. Though I probably shouldn't have given the context.

My bad.

 

[tamtam402]

If you want to be fancy, the power series expansion of sinx is:

sinx = x - x³/6 + ...

It follows that sin(1/x) = 1/x - (1/x)³/6 + ...

and x²sin(1/x) = x - (1/x)/6 + ...

so, lim x -> ∞ x²sin(1/x) = 0