How to prove limits using epsilon-delta definition? I find all steps confusing even for simple examples. As far as i understand, if we have certain positive number epsilon, then we need to prove we can find such delta, such that for all x that satisfy certain inequality with delta, also satisfy inequality with epsilon.
#Epsilon-delta proofs
unreal sequoia
pale cliffBOT
golden shadow
That's pretty much what the definition says, yes.
It's not an easy concept to grasp at first, and the way to get your inequalities varies quite a lot depending on the function, but as you see more examples and work through more exercises, you will start to see what you need to do.
Usually you should do this in two steps : a scratchwork step to bound |f(x) - L| by |x-a|, and then an actual proof step where you pick your delta accordingly.
So for the scratchwork you would start with $|f(x) - L|$ and write a sequence of inequalities in order to get something like $|f(x) - L| < \ldots < (\text{something in terms of $|x-a|)$}$.\
Then if whatever you get on the last step is simple enough, you should be able to go from there and choose $\delta$ such that this expression is less than $\epsilon$.
final gladeBOT
Azyrashacorki
golden shadow
Once you've chosen that you can go back and write the actual proof in the language of the definition, like : Given $\epsilon > 0$, pick $\delta = \text{(what you found in your scratchwork)}$. Then, if $|x-a| < \delta$, $$|f(x) - L| < \ldots < \epsilon.$$
final gladeBOT
Azyrashacorki
golden shadow
Is there any example in particular where you weren't getting the steps?
unreal sequoia
Yes, even in simple limit like x^2, when x approaches 2. So we need to prove that limit is 4
golden shadow
Indeed, so in this case you would start with $|x^2 - 4|$ and try to write a chain of inequalities where $|x-2|$ appears.
final gladeBOT
Azyrashacorki
golden shadow
Okay well this is good, we have $|x-2|$ in there, but we need to deal with $|x+2|$.
Ideally we want $|x+2|$ to not be too big, since we would like to write something like $|x-2 | < \frac{\epsilon}{|x+2|}$.
final gladeBOT
Azyrashacorki
golden shadow
**Azyrashacorki**
There is a trick that is used often in those cases, which essentially relies on seeing that what we're only in interested in the case where x is close to 2
unreal sequoia
We just write epsilon/4?
golden shadow
So we may assume delta is, say, less than 1. This will ensure that |x-2| < 1
unreal sequoia
Oh ok, i just guessed randomly earlier
golden shadow
So if we assume that |x-2| < 1, we get that -1<x-2 < 1 and so 3 < x+2 < 5. This means that |x+2| < 5
unreal sequoia
So we just multiply epsilon by 5?
golden shadow
Well now you would have $|x^2-4| = |x+2|\cdot |x-2| < 5|x-2| < \epsilon$, so you may choose $\delta = \frac{\epsilon}{5}$.
final gladeBOT
Azyrashacorki
golden shadow
The only thing we need to keep in mind is that since we assumed that $|x-2| < 1$, we need $\delta \le 1$ as well, so you may pick $\delta$ to be the least between 1 and $\frac{\epsilon}{5}$.
final gladeBOT
Azyrashacorki
golden shadow
So then your choice would be $\delta = \min{1, \frac{\epsilon}{5}}$.
final gladeBOT
Azyrashacorki
golden shadow
Yes, 1 would then be the least anyways
unreal sequoia
I get confused about this part: Why when we have |x-2| < delta and |x-2| < k * epsilon, where k is random constant, we say that if delta = k * epsilon, then we have proved that such delta exists?
Why we even need left sides equal?
golden shadow
Well the point is you get your chain of inequality, and whatever you end up with has to be bounded by epsilon
So for $x^2$ above, you would get your final proof as : Given $\epsilon > 0$, pick $\delta = \min\left{1, \frac{\epsilon}{5} \right}$. Then $$|x^2 - 4| = |x+2| \cdot |x-2| < 5 |x-2| < 5\frac{\epsilon}{5} = \epsilon.$$
final gladeBOT
Azyrashacorki
unreal sequoia
Oh so, if we have |x-a| < K, where K is epsilon and constant combination and |x-a| < delta, then if we let delta equal K, then for all x values that satisfy delta inequality, will also satisfy epsilon inequality, because they are equal!?
However, how do we know there will be at least 1 x solution to delta inequality always?
golden shadow
Well if $K\epsilon$ is a number you can just set $\delta = K\epsilon$.
final gladeBOT
Azyrashacorki
golden shadow
In general it may not be just in terms of a constant, you may have to pick epsilon^2 or more complicated expressions, so long as this means that your last inequality is < epsilon
unreal sequoia
So is my last sentence - 1 is correct?
The statement was: If we let delta = epsilon*C, then we will have proved everything, because last inequality left sides are equal?
zenith plume
If we let $\delta = \epsilon \cdot C$, then we have proved the limit exists, because we have shown that the condition $|x-a| < \delta$ guarantees that $|f(x) - L| < \epsilon$ through a chain of inequalities.
final gladeBOT
Canvas123
unreal sequoia
Ok, but how to know if |x-a|<delta has at least 1 x solution when we set delta = epsilon * C?
golden shadow
Ok, but how to know if |x-a|<delta has at least 1 x solution when we set delta =...
If you’re evaluating at a point in the domain and delta is nonzero there’s always an interval of size delta around that point
unreal sequoia
If you’re evaluating at a point in the domain and delta is nonzero there’s alway...
And we know that delta will not be ever 0, because epsilon is positive?
Therefore it always works?
golden shadow
Provided what you get in your last inequality isn't negative, yes.
unreal sequoia
Last step i do not understand from proof x^2 you showed earlier is the following: why are we allowed to assume |x-2| < 1 and why delta then needs to be less than or equal than 1?
So we can assume anything as long as we can find such number delta, that satisfies all?
golden shadow
Delta will be small anyways, this is just a simplifying assumption to help estimate |x+2|
We're allowed to do it because if epsilon/5 is larger, we can just choose a smaller delta if we want to and the working will still work.
unreal sequoia
Ok, can you give me tricky problems, where i have to find delta, such that it satisfies all?
For example, if we had x^2+2x+3, how can we find limit if x converges to 2? We can not factorize it...
golden shadow
Where do you want to evaluate this limit? Before you prove a limit is some value you need to know what limit you're computing and what value you expect it to be.
a=-1?
unreal sequoia
no a = 2
golden shadow
So what limit do you think will work?
unreal sequoia
11
golden shadow
Ok, so you need to start with $|x^2+2x+3 - 11|$
final gladeBOT
Azyrashacorki
golden shadow
Work out a chain of inequalities with |x-2| in it
unreal sequoia
Oh nvm we can factorize it now... How about example x^2+2x+13 so that we get x^2+2x+2 when 11 is subtracted and we can not factorize it?
golden shadow
For polynomials you'll always be able to factorize it since you know there's a root at x=a
unreal sequoia
Oh yes, because now it will not be 11...
unreal sequoia
For polynomials you'll always be able to factorize it since you know there's a r...
Yes i get it
So we factorize and go through the same steps as for x^2 example?
golden shadow
Yes, it should look similar with a simplifying assumption
unreal sequoia
Alright, are there trickier problems?
So we can prove any limit with epsilon-delta definition, but the larger and harder problem is, the messier and larger proof becomes so we just use continuous function definition and plug numbers in?
golden shadow
Yes, it should look similar with a simplifying assumption
Note that the choice of delta < 1 in this case is arbitrary. You could've chosen any number and it would've bounded |x+2| (or a similar factor for a proof for polynomials).
1 is nice to work with though
unreal sequoia
Note that the choice of delta < 1 in this case is arbitrary. You could've chosen...
Ok
golden shadow
So we can prove any limit with epsilon-delta definition, but the larger and hard...
The initial plugging in is just because the functions we're considering are continuous, so "guessing" the limit is easy.
unreal sequoia
But can we prove any limit using delta-epsilon definition?
golden shadow
Yes
unreal sequoia
Sounds very interesting, but not convenient at all, because proofs i think get very large if we have hard limit equation
golden shadow
You can try with $\lim_{x\to 9}( \sqrt{x}+2)$
final gladeBOT
Azyrashacorki
unreal sequoia
Alright
golden shadow
Again it's continuous so you can guess what the limit is
But it's a similar approach
unreal sequoia
I found this example quite confusing, but i guess answer is d = e^2 + 6e
Do not know whether it is correct or not
unreal sequoia
But it's a similar approach
So what is the correct answer if you know?
zenith plume
So what is the correct answer if you know?
your guess is way off. $\delta$ should be something like $ \frac{\epsilon}{\sqrt{9}+2} $. just use the conjugate and bound $|x-a|$ by 1 again if you want to keep it simple... this shouldn't take long once u get how conjugates work
final gladeBOT
Canvas123
unreal sequoia
Oh ok
zenith plume
why would $\delta$ be larger than epsilon if the function is growing?
final gladeBOT
Canvas123
unreal sequoia
I do not know, i guess i still do not get how these proofs work...
All i want to understand is: How to learn how to prove any limit using epsilon-delta definition?
zenith plume
All i want to understand is: How to learn how to prove any limit using epsilon-d...
try rationalizing that square root expression instead of guessing numbers at this point or ur gonna be here all night
you don't "learn" it like a formula, you just practice bounding the error term
unreal sequoia
As far as i know, we need to get left sides of inequality equal, where left side has function.
But how to do it in hard examples like (6+4x)/(x^2+1) where x converges to -3?
zenith plume
you don’t need equality
u literally only care about finding an upper bound for $|f(x)-L|$ in terms of some constant times $(|k+3|) \dots$ with rational functions you use that $\delta < 1/(\text{something})$ trick to keep the denominator from blowing up or hitting zero...
final gladeBOT
Canvas123
zenith plume
once it's bounded away by a minimum value then everything just collapses back into polynomial-style bounding
unreal sequoia
Is there another method to find such delta without getting left sides of inequality equal?
Because getting sides equal seems very hard
golden shadow
The problem with what you'd done above is that you need to tweak the delta you got slightly so it works (which it can work), but then once that's done actually going back and proving that |sqrt(x) - 3| < epsilon from it is quite hard.
What do you mean by getting sides equal?
What this helper was suggesting was to write $$\abs{\sqrt{x}-3} = \abs{(\sqrt{x}-3) \cdot \frac{\sqrt{x}+3}{\sqrt{x}+3}}}= \abs{\frac{x-9}{\sqrt{x}+3}}.$$
final gladeBOT
Azyrashacorki
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
golden shadow
Then it becomes a similar problem of having to bound sqrt(x) + 3 by forcing an extra bound on delta.
But again it would be good to clarify what you mean by getting the left sides equal
unreal sequoia
If we have 2 inequalities with epsilon on the right side in 1st inequality and delta on the right in 2nd inequality, then making left sides equal, we can just let delta = epsilon and problem solved?
golden shadow
What you're writing seems like you're asking to choose $\delta = \epsilon$ in any case since $|x-a| < \delta$ and $|f(x) - L| < \epsilon.$
final gladeBOT
Azyrashacorki
golden shadow
The goal is always the same : you want to be able to bound $|f(x) - L|$ by some expression of $|x-a|$ in order to be able to choose an appropriate $\delta$ that will make $|f(x) - L| < \epsilon.$
final gladeBOT
Azyrashacorki
unreal sequoia
Yes, but how to practically do it?
golden shadow
There isn't a universal method.
You've seen that for polynomials there's a certain running theme.
With square roots, it's generally a good idea to conjugate.
unreal sequoia
Alright, but we are still always left with squareroots, are not we? How to get rid of them completely?
golden shadow
**Azyrashacorki**
Compile Error! Click the <:errors:504266556054962196> reaction...
From here you can see you have $|x-9|$ in the numerator, which is good since we're able to bound that explicitly by delta. From here there's a few things you can do. You can bound delta, but you can also see that $\frac{1}{\sqrt{x} + 3} \le \frac{1}{3}$ for all $x$.
final gladeBOT
Azyrashacorki
golden shadow
So then you would find that $\abs{\sqrt{x} - 3} = \frac{\abs{x-9}}{\abs{x+3}} \le \frac{\abs{x-9}}{3}$. At this point we want to bound this by $\epsilon$.
final gladeBOT
Azyrashacorki
golden shadow
So we need to get $\delta$ such that if $\abs{x-9} < \delta$, $\frac{\abs{x-9}}{3} < \epsilon.$
And indeed that would require that $\frac{\delta}{3} < \epsilon$, so $\delta = 3\epsilon$ would work
In this case clearly picking $\delta = \epsilon$ would also work
unreal sequoia
Hmm, very nice proof. Thanks for explaining it.
golden shadow
So the whole way this works is again by working with |f(x) - L| to get a bound on it which depends on |x-a|, since this is what we can "control".
unreal sequoia
What if we had equation like (6+4x)/(x^2+1) where x converges to -3. How to make it equal to x+3? It seems impossible... Is there a way?
golden shadow
Well the limit in this case should be -3/5
I would assume writing things down and setting a common denominator that you could get x+3 to appear in there
unreal sequoia
Oh i actually found a way i think
We let numerator be 4(x+3)-6 first
Then we see that 1/(x^2+1)< 1 so we see that (4(x+3)-6)/(x^2+1) < 4(x+3)-6 < e
And x+3 < (e+6)/4
So delta is (e+6)/4
Is it correct?
golden shadow
You can write the numerator like that but you still need to subtract the limit.
Remember what you want to bound if |f(x) - L|, not f(x)
golden shadow
You will get a similar situation where you need to bound some term which works nicely if you force an extra bound on delta
unreal sequoia
Yes
unreal sequoia
Wait, why delta needed to be less than or equal to 1 if we assumed that |x-2| < 1?
It can also be larger than 1 and it will work, will it not?
golden shadow
Wait, why delta needed to be less than or equal to 1 if we assumed that |x-2| < ...
For which problem?
golden shadow
It may work for some values but you’re not guaranteed the bound on |x+2| works then
In particular, when epsilon > 5, you lose your bound on |x+2|.
You can see it on here.
The goal for delta to be right is that for any epsilon you pick, the part of the graph of f which is the image of f in the interval (a-delta, a+delta) has to be in the horizontal strip (L-epsilon, L+epsilon). I've drawn this specific part of the graph in red. The goal is that as you change epsilon, this red part of the graph stays within the blue band.
You can check that when epsilon > 6 this fails in this case if you don't bound delta by 1 as well : the red part of the line escapes the blue band.
unreal sequoia
Oh so if we assumed |x-2| < 1 and based on this fact found delta in terms of epsilon, then if we chose delta larger than 1, then it would maybe not work out, because proof only works if |x-2| < 1 or delta is equal or smaller than 1?
golden shadow
Yes, because we assumed |x-2| < 1 when bounding |x+2|
This choice is arbitrary, you could'be similarly chosen |x-2| < 5 and found some corresponding delta for what.
You would've gotten delta = min(epsilon/9, 5).
unreal sequoia
Alright, i think i understand it now
But of course, if epsilon/9 is larger than 5, then setting delta equal to epsilon/9 also works?
golden shadow
Well no
Since then |x+2| isn't less than 9
And your bound fails.
So if epsilon/9 > 5, you can just "stop" delta from growing and leave it at 5
unreal sequoia
Oh ok