Show that limit is equal to zero

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In summary, the conversation discusses the calculation of a limit involving the integral of a function over a small ball and the use of spherical or polar coordinates. The speaker also points out a mistake in the calculation and suggests using the fact that the integral of the function over the boundary of the small ball tends to 0 as the radius of the ball approaches 0.
  • #1
evinda
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Hello! (Wave)Suppose that $B_{\epsilon}=\{ |x- \xi|< \epsilon\}$ and $E(|x- \xi|)=\left\{\begin{matrix}
\frac{1}{(2-n)w_n}|x-\xi|^{2-n} &, n \geq 3 \\ \\
\frac{1}{2 \pi} \ln{|x-\xi|} &, n=2
\end{matrix}\right.$

So when $ |x- \xi|=\epsilon $ then

$$E(|x- \xi|)=\left\{\begin{matrix}
\frac{1}{(2-n)w_n}\epsilon^{2-n} &, n \geq 3 \\ \\
\frac{1}{2 \pi} \ln{\epsilon} &, n=2
\end{matrix}\right.$$I want to show that $\lim_{\epsilon \to 0} \int_{\partial{B_{\epsilon}}} E(|x- \xi|) f(\xi)ds=0$ for any continuous and bounded $f$.

There is a hint that we should check the integral $\int_{|x|=\epsilon} E(|x|) ds=0$ by using spherical $(n \geq 3)$ or polar $(n=2)$ coordinates.So, suppose that $n \geq 3$.

Then I thought the following:

$\int_{|x|=\epsilon} E(|x|) ds=\int_{|x|=\epsilon} \frac{1}{(2-n) w_n} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} \epsilon^{2-n} ds=\frac{\epsilon^{2-n}}{(2-n) w_n} 2 \pi \epsilon^{n-1}=\frac{2 \pi \epsilon}{(2-n) w_n} \to 0 \text{ while } \epsilon \to 0$.

But I didn't use spherical coordinates. Have I done something wrong? (Thinking)
 
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  • #2
evinda said:
So, suppose that $n \geq 3$.

Then I thought the following:

$\int_{|x|=\epsilon} E(|x|) ds=\int_{|x|=\epsilon} \frac{1}{(2-n) w_n} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} \epsilon^{2-n} ds=\frac{\epsilon^{2-n}}{(2-n) w_n} 2 \pi \epsilon^{n-1}=\frac{2 \pi \epsilon}{(2-n) w_n} \to 0 \text{ while } \epsilon \to 0$.

But I didn't use spherical coordinates. Have I done something wrong? (Thinking)

Hey evinda!

Well... there is something wrong... :eek:

It should be:
$$\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} \epsilon^{2-n} ds=\frac{\epsilon^{2-n}}{(2-n) w_n} w_n \epsilon^{n-1}$$
Otherwise I think it's all good! (Happy)
 
  • #3
I see... So like that we show that $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} E(|x|) ds=0$.

How do we use this in order to calculate the limit $\lim_{\epsilon \to 0} \int_{\partial{B_{\epsilon}}} E(|x-\xi|) f(\xi) ds$ ?
 
  • #4
evinda said:
I see... So like that we show that $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} E(|x|) ds=0$.

How do we use this in order to calculate the limit $\lim_{\epsilon \to 0} \int_{\partial{B_{\epsilon}}} E(|x-\xi|) f(\xi) ds$ ?

We have:
$$0 \le \left| \int_{\partial{B_{\epsilon}(x)}} E(|x-\xi|) f(\xi) ds \right| \le \sup_{\xi\in \partial B_\epsilon(x)} |E(|x-\xi|) | \cdot \sup_{\xi\in \partial B_\epsilon(x)} |f(\xi)| \cdot \int_{\partial{B_{\epsilon}(x)}} ds
$$
don't we? (Wondering)
 
  • #5
I like Serena said:
We have:
$$0 \le \left| \int_{\partial{B_{\epsilon}(x)}} E(|x-\xi|) f(\xi) ds \right| \le \sup_{\xi\in \partial B_\epsilon(x)} |E(|x-\xi|) | \cdot \sup_{\xi\in \partial B_\epsilon(x)} |f(\xi)| \cdot \int_{\partial{B_{\epsilon}(x)}} ds
$$
don't we? (Wondering)

Yes. So don't we use the fact that $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} E(|x|) ds=0$ ?

When $\xi \in \partial{B_{\epsilon}(x)}$ do we have that $|x-\xi|=\epsilon$ ?

If so then $\sup_{\xi \in \partial{B_{\epsilon}(x)}} |E(|x-\xi|)|=\frac{1}{(2-n) w_n} \epsilon^{2-n}$ and $\int_{\partial{B_{\epsilon}(x)}} ds=w_n \epsilon^{n-1}$ . Right?
 
  • #6
evinda said:
Yes. So don't we use the fact that $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} E(|x|) ds=0$ ?

When $\xi \in \partial{B_{\epsilon}(x)}$ do we have that $|x-\xi|=\epsilon$ ?

If so then $\sup_{\xi \in \partial{B_{\epsilon}(x)}} |E(|x-\xi|)|=\frac{1}{(2-n) w_n} \epsilon^{2-n}$ and $\int_{\partial{B_{\epsilon}(x)}} ds=w_n \epsilon^{n-1}$ . Right?

Right! (Nod)

And $f$ is given to be bounded, so $\displaystyle\sup_{\xi \in \partial{B_{\epsilon}(x)}} |f(\xi)| = M$ for some $M$.
 
  • #7
Nice... Thank you! (Smile)
 

FAQ: Show that limit is equal to zero

What does it mean to show that a limit is equal to zero?

Showing that a limit is equal to zero means that as the independent variable approaches a certain value, the dependent variable approaches zero. In other words, the function gets closer and closer to the value of zero as the input gets closer to a specific value.

How do you show that a limit is equal to zero?

To show that a limit is equal to zero, you can use various methods such as direct substitution, factoring, or using algebraic manipulation. You can also use the squeeze theorem or the definition of a limit to prove that a limit equals zero.

Why is it important to show that a limit is equal to zero?

Showing that a limit is equal to zero is important because it helps us understand the behavior of a function near a specific point. It also allows us to evaluate limits of more complex functions and use them in applications such as optimization and curve sketching.

What are some common mistakes when trying to show that a limit is equal to zero?

Some common mistakes when trying to show that a limit is equal to zero include not considering the possibility of a removable discontinuity, using incorrect algebraic manipulations, and not checking the left and right-hand limits separately. It is also important to check for any restrictions on the domain of the function.

Can a limit be equal to zero at a point but the function not be defined at that point?

Yes, a limit can be equal to zero at a point but the function not be defined at that point. This can occur when there is a removable discontinuity or a jump discontinuity at that point. In these cases, the limit still exists and is equal to zero, but the function is not defined at that specific point.

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