Is calculating the bounds based on one - sided limit like post #24?
But in post#15, the delta you take is ##|\sqrt[3]{8+\epsilon}-2|##, not the negative one?
By minimum, do you mean ##\delta=\text{min}~(1, \frac{\epsilon}{4})##?
Thanks
Ah my bad
Oh ok, but how about something like this?
That example is from textbook and there is part 2 of the working (to show that this ##\delta## works). Is this part not needed and I can just stop at part 1 after stating ##\delta=\frac{\epsilon}{4}##?
I am sorry I don't get this. In post...
I am trying to but sorry I don't have idea how to continue without choosing a certain value for ##\delta##. What I practice up until now is the one which always choose the value of ##\delta##.
If not choosing any value for ##\delta##, what is the continuation after ##\delta=\text{min}...
Then how to continue the proof?
Let:
##\delta_1=\text{min}\{\sqrt[3] {8-\epsilon}-2, \sqrt[3] {8+\epsilon}-2\}##
##\delta_2=\text{min}\{\sqrt{4-\epsilon}-2, \sqrt {4+\epsilon}-2\}##
##\delta=\text{min} \{\delta_1, \delta_2 \}##
For left limit: ##2-\delta<x<2##. How to change this into ##|x^3...
Wait, I don't need to choose which one is the minimum? But then when I want to prove it, what should I put as ##delta##?
Oh I have to take one common ##\delta## and use it for left and right side limit?
Thanks
For left side limit ##(x\to 2^{-})##, the ##\delta=|\sqrt[3] {8+\epsilon}-2|##
For right side limit, ##(x\to 2^{+})##, the ##\delta=|\sqrt{4+\epsilon}-2|##
Is this what you mean?
Yeah I am having difficulty getting the ##|x-2|## to complete the proof. I am trying to use ##a-\delta<x<a## for...
How to continue the working to complete the proof?
For left side limit, let ##\delta=|\sqrt[3] {8+\epsilon}-2|##. Taking ##\delta>0## and ##0<\epsilon<8## , then ##\delta=|\sqrt[3] {8+\epsilon}-2|=\sqrt[3] {8+\epsilon}-2##
So:
$$2-(\sqrt[3] {8+\epsilon}-2)<x<2$$
$$4-\sqrt[3] {8+\epsilon}<x<2$$...
Sorry I don't understand. Isn't ##f## defined in the question?
For ##x\to 2^{-}, f(x) = x^3 +2## and for ##x\to 2^{+}, f(x)=x^2 + 6##. Is this what you mean?
Ok so the the right side limit, I also assume ##0<\epsilon<4## so the minimum would be ##|\sqrt{4+\epsilon}-2|##
Is it ok to have two...
I am actually asking whether it is ##-\frac{\mu_o abN}{2L}## or ##-\frac{\mu_o abN^2}{2L}## but I think now I know why I am wrong.
I was mistaken when taking which object to consider. I thought I need to consider the number of turns of solenoid. It turns out I need to consider the loop...
I mean why for the flux it is not ##\frac{\mu_o N^2 I}{L}##. I saw something similar to this when calculating self inductance of solenoid; I need to multiply the flux to get total flux linkage so the self inductance can be ##\frac{\mu_o N^2 \pi r^2}{L}##
I understand
The answer key is:
a) ##\frac{\mu_o \pi nar_{1}^{2}}{R}##
b) ##-\frac{\mu_o abn}{2}##
My questions:
1) Why is there no ##\cos \theta## in answer (a)?
$$\text{induced current}=\frac{\varepsilon}{R}$$
When calculating induced emf ##\varepsilon=-\frac{d\Phi}{dt}=-\frac{d(\vec B . \vec A)}{dt}##...