@@marcusholloway1633 the epsilon delta definition assures us I can always find a delta ‘for any ε’. ε is just a stand-in for the idea of ‘any number’. We want to show that for ‘any ε’ we can find a delta for the limit we want to prove. Let's say we have a value ε, and we are struggling to find the delta. Now we know we have two other limits that exist. So I know that for any number, I can always find a delta for those cases. Let’s say the number given just happens to be ε / 2. Then since the limit exists, I can find the corresponding delta, which helps me complete the proof. Similarly, I can also find a corresponding delta for ε / 3, 5ε, or anything else, because the idea is I can find a delta for ‘any number’. In the proof, I used ε / 2 because it conveniently helps me establish the final proof result.
@@marcusholloway1633 yes you can. The final statement you’ll get is ‘for any 2 epsilon’, you can always find a delta. Since epsilon is arbitrary, 2 epsilon is too, so it does satisfy our purpose, but of course presentation wise this would look a bit messy. The point is there’s no need to fuss over the form of the term ‘epsilon’. As long as we can show we can find a delta for any arbitrary positive real number, then the limit is established.
Wow! your explanation is so clear and concise, you didn’t leave anything behind! Very underrated!
Thanks :D not a lot of people appreciate the Epsilon-delta proofs and I'm glad you found the video useful
Easiest to follow explanation I was able to find. Thanks so much.
Very clever and clear way of build this demostration ... even better than many calculus books ... nice
Can you explain the ε/2 in more depth plz
@@marcusholloway1633 what would you like to know?
@@HayashiManabu What is the reason why you can write ε/2 instead of ε?
@@marcusholloway1633 the epsilon delta definition assures us I can always find a delta ‘for any ε’. ε is just a stand-in for the idea of ‘any number’.
We want to show that for ‘any ε’ we can find a delta for the limit we want to prove. Let's say we have a value ε, and we are struggling to find the delta.
Now we know we have two other limits that exist. So I know that for any number, I can always find a delta for those cases. Let’s say the number given just happens to be ε / 2. Then since the limit exists, I can find the corresponding delta, which helps me complete the proof.
Similarly, I can also find a corresponding delta for ε / 3, 5ε, or anything else, because the idea is I can find a delta for ‘any number’. In the proof, I used ε / 2 because it conveniently helps me establish the final proof result.
@@HayashiManabu but why you have to make [f(x)+g(x)]-[L-M]=epsilon in the end.Could it be 2 epsilon? Cuz you say epsilon means any number
@@marcusholloway1633 yes you can. The final statement you’ll get is ‘for any 2 epsilon’, you can always find a delta. Since epsilon is arbitrary, 2 epsilon is too, so it does satisfy our purpose, but of course presentation wise this would look a bit messy.
The point is there’s no need to fuss over the form of the term ‘epsilon’. As long as we can show we can find a delta for any arbitrary positive real number, then the limit is established.
Sir can you please solve problem 4.34(Introduction to quantum mechanics (Griffiths))
Nice one👍
satisfying