Could there be a slight inaccuracy in picking the value for delta? If delta=min{1,epsilon/5}, and assuming the case where 1 is the smaller of the two, then delta=1. But then |x+2|
You are missing something, note |x - 2| < delta, by assumption, so you get the strong inequality right at the beginning, |x^2 - 4| = |x - 2||x + 2| < delta*|x + 2|
Just the way u made an intuitive sense for x-2 to be considered less than one using the x axis number line, and it makes sense, could u show intuitive sense for x+ 5 on the graph? Thanks
It must be some slight error in writing. Technically it should be this. |x-2| |x+2| < epsilon since |x-2| < delta therefore delta * |x+2| = epsilon That being |x+2| is less than something.
I don't get how we know that x is within 1 of 2. It seems to me that you are saying just vaguely that it could be within 1 and then we are just working off of that assumption but what then in the case where x is =3.5 or something like.
I do not understand why ( in part 2, proof proper), you write that |x - 2| . |x+2| < delta.5. I see that " delta.5 " is the product of two " bonds" , also see that |x-2| is true by hypothesis, but by where does |x+2|
that’s how absolute value works, observe, if i told you |x| < 3, then x could only take values from but not including -3 and 3, if i told you x was -4 then put the absolute operator back on the inequality would no longer hold
we have chosen delta to be smaller than 1 and smaller than epsilon over 5, then i dont understand how the minimum of the 2 would be smaller than the 2? wouldnt it rather be equal to one of the 2 numbers than being smaller than the 2? hope someone can clear this up.
Sure go ahead and pick 2/3/sqrt(2), the whole point of these proofs is to show no matter how sufficiently small our delta and epsilon are, we can always find a satisfying condition. We pick delta = 1 out of convenience to get this satisfying condition. (Remember, we can let delta be whatever, and all we wish is to find a corresponding epsilon inequality )
Thank you a lot! For me this was the most helpful video about this topic.
Thanku so much sir 👍👍 love from India.......i see It three 4 times ....and now I can solve any quadratic problem,.
Could there be a slight inaccuracy in picking the value for delta? If delta=min{1,epsilon/5}, and assuming the case where 1 is the smaller of the two, then delta=1. But then |x+2|
You are missing something, note |x - 2| < delta, by assumption, so you get the strong inequality right at the beginning,
|x^2 - 4| = |x - 2||x + 2| < delta*|x + 2|
@@TheMathSorcerer Thanks for clearing that up, friend!
This video really helped me. Thanks!
Excellent!
Thank you so much. I had a very similar question to this in my text book where x->-1 and they said |x-3| 4
The limit-value is 2 lim x->2 2^2=4
It's so abstract, I don't know what this is all about. It's like proving 1+1=2, why does it even need proving.
Some people wrote hundreds of pages to prove 1=2.
I am also thinking that i hope physics has nothing to do with this bullahit proving.....btw i am doing physics
Ur doing mid level math and is wondering why we need prove things???
Ur doing mid level math and is wondering why we need prove things???
Ur doing mid level math and is wondering why we need prove things???
Thank you very much, helped a lot.
Glad to hear that!
@@TheMathSorcerer I never knew how we set those deltas coming out of nowhere lol. I will write scratchwork and proof separately on my exam paper xd
Just the way u made an intuitive sense for x-2 to be considered less than one using the x axis number line, and it makes sense, could u show intuitive sense for x+ 5 on the graph?
Thanks
That is a pretty good video, but there's one thing that I don't understand:
after I have the simplification |x^2-4| = |x-2|*|x+2|
I totally agree! Please sir, have you found an answer between your comment and now?
I think they just assume it would be bigger because it was an input and ouput anyway
It must be some slight error in writing.
Technically it should be this.
|x-2| |x+2| < epsilon
since |x-2| < delta
therefore
delta * |x+2| = epsilon
That being |x+2| is less than something.
Siii, al fin encontré la explicación a este ejercicio
Que bueno!!
I don't get how we know that x is within 1 of 2. It seems to me that you are saying just vaguely that it could be within 1 and then we are just working off of that assumption but what then in the case where x is =3.5 or something like.
why do we want delta*|x + 2| < 5*delta to be less than Epislon? 5:29
If something is *less* than delta why can we replace it *with* delta. Where you replaced (x-2) with delta at the start...
Sir @03 :00 why you took 1 and 3 as close to 2 why not 1.9 and 2.I or something similar like that.
you could have chosen those for sure and it would work yes,no reason, just 1 is simple, your example works also
I do not understand why ( in part 2, proof proper), you write that |x - 2| . |x+2| < delta.5. I see that " delta.5 " is the product of two " bonds" , also see that |x-2| is true by hypothesis, but by where does |x+2|
Correction, line 2 : " I know that | x-2| < delta is true by hypothesis...."
@@raylittlerock3940 you can edit the comment my man
With delta = 1 we get (x-2)0, we know that (x-2) = |x-2|
A property says "if |x|0 then -a
Crystal clear!
I found that, delta = 1 works for every E > = 5. I have to find a delta for E < 5, but I dont know how
could you expain why at 4:13 we can make the assumtion that we can put -1 in -1
that’s how absolute value works, observe, if i told you |x| < 3, then x could only take values from but not including -3 and 3, if i told you x was -4 then put the absolute operator back on the inequality would no longer hold
What if x---> -2 in this limit
Can I say that since |x-2| < delta, then |x+2| < delta +4, so |x-2| |x+2| < delta (delta +4)?
No, you can say that since |x-2| < delta, then |x-2| + 4 < delta + 4, but |x-2| + 4 is not the same as |x+2| (they only agree for x >= 2).
we have chosen delta to be smaller than 1 and smaller than epsilon over 5, then i dont understand how the minimum of the 2 would be smaller than the 2? wouldnt it rather be equal to one of the 2 numbers than being smaller than the 2? hope someone can clear this up.
I think he said it wrong, it should be equal to
hey bro can I pick some number other than 1, say 2/3/sqrt2
yes
Sure go ahead and pick 2/3/sqrt(2), the whole point of these proofs is to show no matter how sufficiently small our delta and epsilon are, we can always find a satisfying condition. We pick delta = 1 out of convenience to get this satisfying condition. (Remember, we can let delta be whatever, and all we wish is to find a corresponding epsilon inequality )
excellent
Fire emojis
rushed explanation, if youre teaching it to someone who doesnt know make sure its a clear one pls