Will epsilon delta test fail if curve changed direction?

[AutumnWater]

Will epsilon delta test fail if curve changed direction within the +/- delta of the limit point?

Is there a scenario where no matter how small we pick delta to be, the frequency of the graph changing directions is always going to be a higher than delta's distance? In that scenario the limit does not exist?

 

[axmls]

What do you think? What quantities are important in the epsilon-delta definition?

 

[AutumnWater]

So if we have sin(kx) as k->infinity, I'm thinking in that scenario the graph just becomes a vertical line between [-1 1] all over the place, how are we supposed to find limit then?

 

[axmls]

What is $x$ in this situation? This looks more like a two-variable limit. In other words, what is $x$?

Consider the simpler case $\lim_{x \to \infty} \sin(x)$. What is the epsilon-delta definition? Well, we must use a slightly different definition for infinite limits.

We say that $\lim_{x \to \infty} f(x) = L$ if, given any $\epsilon > 0$, there is some $N$ such that if $x > N$, then $|f(x) - L| < \epsilon$.

So let's apply that to $\sin(x)$.

Let's say I give you some small value $\epsilon$. Let's assume (for now) that $\lim_{x \to \infty} \sin(x) = L$ for some real number $L$ (which is obviously between $-1$ and $1$. Can you find some real number $N$ so that the distance between $\sin(x)$ and $L$ is less than that value of $\epsilon$ whenever $x > N$?

 

[AutumnWater]

What do you think? What quantities are important in the epsilon-delta definition?

Well epsilon and delta, rise over run in the 1d sense, so if rise was negated by the change of direction, then the ratio becomes inaccurate? that was my intuition.

 

[axmls]

Well epsilon and delta, rise over run in the 1d sense, so if rise was negated by the change of direction, then the ratio becomes inaccurate? that was my intuition.

You may need to review the $\epsilon-\delta$ definition. I'll repeat it here in the case of a finite limit.

We say that $\lim_{x \to c} f(x) = L$ if, for every $\epsilon > 0$, there is some $\delta$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x - c| < \delta$.

In other words, assume I'm picking the epsilons. Let's say I pick $\epsilon = 0.1$. Can you find a number $\delta$ so that, whenever the distance between $x$ and $c$ (the point we're approaching) is less than $\delta$, then the distance between $f(x)$ and $L$ (the limit) is always less than $\epsilon$?

If you can always find that $\delta$ no matter how small of an $\epsilon$ I give you, then the limit exists, and it is equal to $L$.

 

[AutumnWater]

I remember this definition, what I have trouble understanding from the definition is how the:

"you can get f(x) as close as you want (within epsilon) to L by making x sufficiently close to c"
translated into:
"x within delta of c implying f(x) within epsilon of L"

and furthermore, when I saw a video of such a proof, the curve being drawn is heading in a straight direction between (c,L) and (c+/-delta, L+/-epsilon) so I'm questioning whether, if a curve changes direction so frequently to the point where no delta can be as small as the rate at which the curve changes directions, then the limit shouldn't exist?

 

[AutumnWater]

What is $x$ in this situation? This looks more like a two-variable limit. In other words, what is $x$?

Consider the simpler case $\lim_{x \to \infty} \sin(x)$. What is the epsilon-delta definition? Well, we must use a slightly different definition for infinite limits.

We say that $\lim_{x \to \infty} f(x) = L$ if, given any $\epsilon > 0$, there is some $N$ such that if $x > N$, then $|f(x) - L| < \epsilon$.

So let's apply that to $\sin(x)$.

Let's say I give you some small value $\epsilon$. Let's assume (for now) that $\lim_{x \to \infty} \sin(x) = L$ for some real number $L$ (which is obviously between $-1$ and $1$. Can you find some real number $N$ so that the distance between $\sin(x)$ and $L$ is less than that value of $\epsilon$ whenever $x > N$?

There'd be an arbitrary amount of Ns that all fit under f(x) - L < epsilon's condition as f(x) oscilated up and down, so the limit doesn't exist?

 

[axmls]

There'd be an arbitrary amount of Ns that all fit under f(x) - L < epsilon's condition as f(x) oscilated up and down, so the limit doesn't exist?

No, it's not necessarily like that. The definition states that for every epsilon, we can find that $N$. If I gave you $\epsilon = 0.1$ (and let's assume the limit is $0$, for instance), is there some $N$ such that whenever $x > N$ then $\sin(x) < \epsilon$? No! No matter how big we make $N$, we can always find a value of $x$ where $x > N$ such that $\sin(x) > \epsilon$. We essentially can't bound the function to a small interval. For instance, $\sin(x) = 1$ for any $x = 2\pi n + \pi/2$.

Or to put it another way, no matter how big we make $N$, we can't guarantee that the distance between $\sin(x)$ and the limit is within the bounds set by our value of $\epsilon$.

Now, if we talk about $f(x) = \sin(x)/x$, then the limit does actually exist as $x \to \infty$ even though the function oscillates, because we actually can find an $N$ to satisfy our condition given any $\epsilon$.

 

[AutumnWater]

No, it's not necessarily like that. The definition states that for every epsilon, we can find that $N$. If I gave you $\epsilon = 0.1$ (and let's assume the limit is $0$, for instance), is there some $N$ such that whenever $x > N$ then $\sin(x) < \epsilon$? No! No matter how big we make $N$, we can always find a value of $x$ where $x > N$ such that $\sin(x) > \epsilon$. We essentially can't bound the function to a small interval. For instance, $\sin(x) = 1$ for any $x = 2\pi n + \pi/2$.

Or to put it another way, no matter how big we make $N$, we can't guarantee that the distance between $\sin(x)$ and the limit is within the bounds set by our value of $\epsilon$.

Now, if we talk about $f(x) = \sin(x)/x$, then the limit does actually exist as $x \to \infty$ even though the function oscillates, because we actually can find an $N$ to satisfy our condition given any $\epsilon$.

sorry I'm a little lost as to the X being "larger than N", I thought |x-c| < delta, so I was looking at delta here instead of N, how'd you get x > N when x was on the LHS of the < sign with |x-c| < delta? shouldn't it be |x- pi/2| < delta in this case?

 

[axmls]

It depends on if you're talking about infinite limits or not. For infinite limits, we look for the existence of large $N$ that imply $|f(x) - L| < \epsilon$, but when we're looking at the limit as $x \to c$, we're looking for $\delta$ such that $|x - c| < \delta \implies |f(x) - L| < \epsilon$.

In the infinite limit case, which this thread was originally about, we want to show that if $x$ is sufficiently big (that is, there's some $N$ such that $x > N$), then the function in question falls within the bounds decided by our $\epsilon$. It just so happens in this case that there is no such $N$ that satisfies this. The limit does not exist because no matter how large $x$ gets, we can never guarantee the distance between $f(x)$ and $L$ will shrink.

 

[AutumnWater]

It depends on if you're talking about infinite limits or not. For infinite limits, we look for the existence of large $N$ that imply $|f(x) - L| < \epsilon$, but when we're looking at the limit as $x \to c$, we're looking for $\delta$ such that $|x - c| < \delta \implies |f(x) - L| < \epsilon$.

In the infinite limit case, which this thread was originally about, we want to show that if $x$ is sufficiently big (that is, there's some $N$ such that $x > N$), then the function in question falls within the bounds decided by our $\epsilon$. It just so happens in this case that there is no such $N$ that satisfies this. The limit does not exist because no matter how large $x$ gets, we can never guarantee the distance between $f(x)$ and $L$ will shrink.

ok that makes more sense thanks,

what if in sin(kx) where x is approaching c, but k is approaching infinity? do we use N or c here?

we can't use 0< sqrt((x-c)^2 + (k-infinity)^2) < delta here can we?
But if the function we're speaking of is not a periodic one like sine function, and it merely changes direction only once, but the distance between the critical point and its c+delta and c-delta are so small, that it approaches 0, can we still say that using the |x-c|<delta method, that we won't be able to find such a delta?

 

[axmls]

But if the function we're speaking of is not a periodic one like sine function, and it merely changes direction only once, but the distance between the critical point and its c+delta and c-delta are so small, that it approaches 0, can we still say that using the |x-c|<delta method, that we won't be able to find such a delta?

The great thing is that whether or not the limit exists has nothing to do with changing directions. As long as we can get close enough to $c$ (within $(c - \delta, c + \delta)$), and when we do so, it guarantees that $f(x)$ is close enough to $L$ (within $(L - \epsilon, L + \epsilon)$), then the limit exists and is equal to $L$. In your scenario (if I understand correctly), let's imagine we have a function that looks like it's going to touch a point, then when it gets really close, it shoots up, then it gets even closer and shoots back down and doesn't behave erratically after that (and it's still approaching the limit). Then of course the limit exists, because as long as we choose our $\delta$ interval so that it doesn't include that big jump, we can still satisfy the $\epsilon-\delta$ condition.

Things are a bit trickier with multivariable limits, because it depends on what path you take in the plane. In essence, you have to show that the limit is the same regardless of the path you take.

 

[AutumnWater]

What about like this? wouldn't c + delta and c - delta both give L + epsilon instead of having an L - epsilon? This would mean there exists 2 deltas both give the same epsilon, so the delta to epsilon relationship doesn't have to be unique?

 

[axmls]

You're not looking for which epsilon a delta gives you. Someone comes up to you and gives you an epsilon, and you have to find the delta that guarantees the function will be within that epsilon of the limit. In your particular case (I'm assuming you meant to draw the two lines so that it would be continuous except for the jump only at c), L is not the limit, because the two lines do not approach the dot in the center. They approach the gap. It doesn't matter what actually happens at x = c.

 

[AutumnWater]

Alright gotcha thanks!

 

[Svein]

For a non-trivial example, try to find out whether sin(1/x) is continuous at 0:
http://demo.activemath.org/ActiveMath2/LeAM_calculusPics/sin%281overx%29_zoom.png?lang=en

 

[AutumnWater]

sin(1/0) is undefined, but limit of sin(1/x) as x approach 0, should be 0, since f(a) does not equal lim x->a f(x), it's not continuous?
to prove the limit:
we need to find a delta such that: |x| < delta implies |sin(1/x)| < epsilon;
since sin(1/x) is always between -1, to 1, |sin(1/x)| <= 1

maybe squeeze theorem can be used here, let me think...

can we take arcsin on both sides of |sin(1/x)| < epsilon? |1/x| < arcsin(epsilon), 1/|arcsin(epsilon)| < |x| so if we set delta = 1/|arcsin(epsilon)|? then |x| < 1/|arcsin(epsilon)| => 1/|x| < |arcsin(epsilon)| so sin(1/|x|)<epsilon?

Ok I messed up...it doesn't exist :(

Can I interpret the non existence as: Because the graph is a straight vertical line at 0, y takes on all values between [-1, 1] so when being handed out an epsilon, the L+epsilon or L - epsilon difference VARIES and we can't find any delta that can accommodate multiple values of epsilon+L or L-epsilon so the limit doesn't exist?

E.G. at x = 0, y can be both 1 and 0 and -1, so if we assumed L = 0, then when |f(x)-L| < epsilon, the f(x) can take on an infinite many number of values such as 1, 0.99999, 0.999998, so on, and there is no single delta that can make epsilon work for all of those f(x)s, that's why it doesn't exist?

But suppose I wasn't given the graph, but only the equation sin(1/x) = y, how do I manipulate the algebra to disprove the limit's existence? lim x->0+ and lim x->0- being different?

Generally speaking, if we have sin(f(x)) = y
Can we just consider the limit of f(x) first, and if that limit exists as L, then just pinpoint a limit of sin(L) for sin(f(x))? since lim x->0- 1/x does not equal lim x->0+ 1/x, then 1/x doesn't have a limit at 0, so sin(1/x) won't either?

I just tested this on sin(sin(x)/x)'s limit at x->0, and it worked as sin(1).

 

[Svein]

Can I interpret the non existence as: Because the graph is a straight vertical line at 0, y takes on all values between [-1, 1] so when being handed out an epsilon, the L+epsilon or L - epsilon difference VARIES and we can't find any delta that can accommodate multiple values of epsilon+L or L-epsilon so the limit doesn't exist?

You are close. In reality, every point on the line (0,-1) to (0, 1) is a cluster point, that is, there are infinitely many points on sin(1/x) close to it.

 

[axmls]

Notice that the fact that $\lim_{x \to 0} \sin(1/x)$ does not exist is the exact same thing as the fact that $\lim_{x \to \infty} \sin(x)$ does not exist, and you can see this if you make the substitution $x \to 1/x$.