Understanding the Epsilon Delta Definition of a Limit

[WhyNerfZed]

Hi

I'm new to limits and calculus in general. Our professor told us there existed some rigorous proof for a limit, but it was "beyond the scope of the course". All we needed to know about a limit was that (1)$$\lim_{x\to a} f(x)$$ is true iff when x approaches a from both directions p(x) approaches L. But i hate using something without being able to properly define it, so i looked up the epsilon-delta definition of a limit. After looking at it for some time, I think I am close to understanding it, but there are still something about it that just don't click in my head(it's so abstract to me). In this thread I'm going try lay out my understanding of the epsilon delta definition, and how it is equivalent to the f(x) → L when x → a^ definition. I hope you can correct me if/when I'm wrong, and help me see how, intuitively, the epsilon-delta definition says the same thing as the other definition. So here goes.

My first issue is how i decipher the general limit expression $$\lim_{x\to a} f(x)$$
So i will try to understand the one sided limit instead $$\lim_{x\to a^-} f(x)$$
How i see it, is that as x approaches a(from the negative direction), f(x) will approach L incrementally. Which implies that that no matter how close L-ε you want to come to L, there exists some δ such that f(a-δ) will evaluate to precisely L-ε. And since as x gets closer to a, f(x) gets closer to L, it must mean that for every ε there exists some x where x > a-δ Λ x < a will evaluate to function values within <L-ε, L>, or in other words L-F(x)< ε. So if all that i have said is true, then if f(x) approaches L when x approaches a, it must mean that ∀ε ∃δ such that if (a-x)<δ ⇒ (L-f(x))< ε ⇔ f(x)→ L when x→a

In essence, what I'm struggling with is how to rephrase the f(x) → L when x → a^^- definition that i understand, to the epsilon delta definition.
I can see that (f(x) → L when x → a^^-) ⇒ any y=L-ε can be achieved by finding an x sufficiently close to a. However the two statements are not equivalent because the last statement does not imply that f(x) → L when x → a^^-. Because it is possible that x<z<a and f(x) is within ε but f(y) is not. So in order for f(x) → L when x → a^^- and the limit to exists , there must exists some x such that, no matter how close ε you want to get to L, f(x) will be that close and every point beyond x and up to a will evaluate to functions as close or closer to a.

Sorry for the wall of text with few, if any, direct questions. Would greatly appreciate any corrections to my post or pointers to how i can understand this better. I can't seem to move on with my school work until i have internalized this definition.

 

[HallsofIvy]

Hi

I'm new to limits and calculus in general. Our professor told us there existed some rigorous proof for a limit, but it was "beyond the scope of the course". All we needed to know about a limit was that (1)$$\lim_{x\to a} f(x)$$ is true iff when x approaches a from both directions p(x) approaches L. But i hate using something without being able to properly define it, so i looked up the epsilon-delta definition of a limit. After looking at it for some time, I think I am close to understanding it, but there are still something about it that just don't click in my head(it's so abstract to me). In this thread I'm going try lay out my understanding of the epsilon delta definition, and how it is equivalent to the f(x) → L when x → a^ definition. I hope you can correct me if/when I'm wrong, and help me see how, intuitively, the epsilon-delta definition says the same thing as the other definition. So here goes.

My first issue is how i decipher the general limit expression $$\lim_{x\to a} f(x)$$
So i will try to understand the one sided limit instead $$\lim_{x\to a^-} f(x)$$
How i see it, is that as x approaches a(from the negative direction), f(x) will approach L incrementally. Which implies that that no matter how close L-ε you want to come to L, there exists some δ such that f(a-δ) will evaluate to precisely L-ε. And since as x gets closer to a, f(x) gets closer to L, it must mean that for every ε there exists some x where x > a-δ Λ x < a will evaluate to function values within <L-ε, L>, or in other words L-F(x)< ε. So if all that i have said is true, then if f(x) approaches L when x approaches a, it must mean that ∀ε ∃δ such that if (a-x)<δ ⇒ (L-f(x))< ε ⇔ f(x)→ L when x→a

In essence, what I'm struggling with is how to rephrase the f(x) → L when x → a^^- definition that i understand, to the epsilon delta definition.
I can see that (f(x) → L when x → a^^-) ⇒ any y=L-ε can be achieved by finding an x sufficiently close to a. However the two statements are not equivalent because the last statement does not imply that f(x) → L when x → a^^-. Because it is possible that x<z<a and f(x) is within ε but f(y) is not.

I am afraid this has me very confused! There is no "y" in anything previous to this. Where did y come from?

So in order for f(x) → L when x → a^^- and the limit to exists , there must exists some x such that, no matter how close ε you want to get to L, f(x) will be that close and every point beyond x and up to a will evaluate to functions as close or closer to a.

Again, the way you have stated this is confusing. There does NOT "exist some x" nor do you want ε to be close to L. You want to be able to show that f(x) can be made "arbitrarily close" to L just by making x "close enough" to a. The purpose of the "$\epsilon$" and "$\delta$" is to make "arbitrarily close" and "close enough" precise: f(x) is "arbitrarily close" to L if |f(x)- L| is less than $\epsilon$ where $\epsilon$ can be arbitrarily chosen and x is "close enough" to a if |x- a| is less than $\delta$.

Sorry for the wall of text with few, if any, direct questions. Would greatly appreciate any corrections to my post or pointers to how i can understand this better. I can't seem to move on with my school work until i have internalized this definition.

 

[Svein]

The definition of $\lim_{x\rightarrow a}f(x)=L$ is: Given an ε>0, if you can find a δ>0 such that $|f(x)-L|<\varepsilon$ whenever $|x-a|<\delta$, then the limit exists and is equal to L.

 

[Mark44]

Hi

I'm new to limits and calculus in general. Our professor told us there existed some rigorous proof for a limit, but it was "beyond the scope of the course". All we needed to know about a limit was that (1)$$\lim_{x\to a} f(x)$$ is true

No. A limit expression such as you have here isn't either "true" or "false." It is a number, provided that the limit exists.

iff when x approaches a from both directions p(x) approaches L. But i hate using something without being able to properly define it, so i looked up the epsilon-delta definition of a limit. After looking at it for some time, I think I am close to understanding it, but there are still something about it that just don't click in my head(it's so abstract to me). In this thread I'm going try lay out my understanding of the epsilon delta definition, and how it is equivalent to the f(x) → L when x → a^ definition. I hope you can correct me if/when I'm wrong, and help me see how, intuitively, the epsilon-delta definition says the same thing as the other definition. So here goes.

My first issue is how i decipher the general limit expression $$\lim_{x\to a} f(x)$$
So i will try to understand the one sided limit instead $$\lim_{x\to a^-} f(x)$$
How i see it, is that as x approaches a(from the negative direction), f(x) will approach L incrementally. Which implies that that no matter how close L-ε you want to come to L, there exists some δ such that f(a-δ) will evaluate to precisely L-ε. And since as x gets closer to a, f(x) gets closer to L, it must mean that for every ε there exists some x where x > a-δ Λ x < a will evaluate to function values within <L-ε, L>, or in other words L-F(x)< ε. So if all that i have said is true, then if f(x) approaches L when x approaches a, it must mean that ∀ε ∃δ such that if (a-x)<δ ⇒ (L-f(x))< ε ⇔ f(x)→ L when x→a

In essence, what I'm struggling with is how to rephrase the f(x) → L when x → a^^- definition that i understand, to the epsilon delta definition.
I can see that (f(x) → L when x → a^^-) ⇒ any y=L-ε can be achieved by finding an x sufficiently close to a. However the two statements are not equivalent because the last statement does not imply that f(x) → L when x → a^^-. Because it is possible that x<z<a and f(x) is within ε but f(y) is not. So in order for f(x) → L when x → a^^- and the limit to exists , there must exists some x such that, no matter how close ε you want to get to L, f(x) will be that close and every point beyond x and up to a will evaluate to functions as close or closer to a.

Sorry for the wall of text with few, if any, direct questions. Would greatly appreciate any corrections to my post or pointers to how i can understand this better. I can't seem to move on with my school work until i have internalized this definition.

 

[micromass]

Epsilon-delta definitions actually have a lot to do with basic physics. It's the same idea of an approximation.

Let's say that you're in a shower. You like your shower to have a temperature of 40°C or approximate. You can regulate your temperature by turning some knob. Of course, you can never get exactly 40°C by turning the knob (you don't have that precision), but you can get close. In fact you can get arbitrary close.

For example, let's say that you like 40°C, but ±5°C is ok too. Then you got a certain number of positions that are ok. In fact, you got an entire range of positions that is ok. The 5°C is called the ε, while the range of positions has to do with the δ.
However, if you're more sensitive and if you like 40°C, but only ±0.5°C is ok. Then there are also some positions of the know that are ok, but there are significantly less positions that are ok.

In general, let's say that you like 40°C with a tolerance of ε°C. Then there is a certain range of positions of the know that are ok. This is the δ-range. The smaller you take ε, the smaller the δ-range will be. But whatever we see, there will always be a δ-range.

So choosing your temperature in the shower is continuous.

Abstractly, you work with functions $f:\mathbb{R}\rightarrow \mathbb{R}$. You should see the domain as some kind of knob in the shower. So we can put the knob on any value we please , but only approximately. The function $f$ regulates the temperature. So let's say we put the knob on $0$, then we will feel temperature $f(0)$. The entire process of epsilon-delta definitions is to get as close as we want to some specific temperature by turning the knob and thus getting within a certain allowable range of the knob. If no matter how close we want to get to the temperature, there is always an allowable interval of the knob to the left and to the right, then the function is continuous.

 

[WhyNerfZed]

Epsilon-delta definitions actually have a lot to do with basic physics. It's the same idea of an approximation.

Let's say that you're in a shower. You like your shower to have a temperature of 40°C or approximate. You can regulate your temperature by turning some knob. Of course, you can never get exactly 40°C by turning the knob (you don't have that precision), but you can get close. In fact you can get arbitrary close.

For example, let's say that you like 40°C, but ±5°C is ok too. Then you got a certain number of positions that are ok. In fact, you got an entire range of positions that is ok. The 5°C is called the ε, while the range of positions has to do with the δ.
However, if you're more sensitive and if you like 40°C, but only ±0.5°C is ok. Then there are also some positions of the know that are ok, but there are significantly less positions that are ok.

In general, let's say that you like 40°C with a tolerance of ε°C. Then there is a certain range of positions of the know that are ok. This is the δ-range. The smaller you take ε, the smaller the δ-range will be. But whatever we see, there will always be a δ-range.

So choosing your temperature in the shower is continuous.

Abstractly, you work with functions $f:\mathbb{R}\rightarrow \mathbb{R}$. You should see the domain as some kind of knob in the shower. So we can put the knob on any value we please , but only approximately. The function $f$ regulates the temperature. So let's say we put the knob on $0$, then we will feel temperature $f(0)$. The entire process of epsilon-delta definitions is to get as close as we want to some specific temperature by turning the knob and thus getting within a certain allowable range of the knob. If no matter how close we want to get to the temperature, there is always an allowable interval of the knob to the left and to the right, then the function is continuous.

Thank you for this intuitive explanation. Would i be right in saying that, given that a limit L exists when x→a, for every ε range around L there exists some δ around a, such that all the points within δ range evaluate to functions within ε. This mean that you can get arbitrarily close to the L(approach L) and if the δ range of points around a is very wide, this mean that every single one of those values are arbitrarily close to L, including when x is arbitrarily close to a. And if the δ is arbitrarily narrow(small) it follows that when x→a f(x) → L.

 

[nuuskur]

Yes, you could say that if you pick a distance from L of some function, then there always exists a neighborhood around the point $a$ whose every function value is strictly closer to L than the distance you specified: $\varepsilon$. Also, we are not concerned about the point $a$ itself only what happens in its vicinity.

On the flipside, were you able to determine an $\varepsilon > 0$ that defies this definition, then you will have proven $\lim_{x\to a} f(x) \neq L$. It's a very handy tool when you are handling limits.