I had been taught the formal definition of a limit with quantifiers. For me, it is very hard to follow and I understand very little of it. I was told that: $$\text{If} \ \lim\limits_{x\to a}f(x)=L, \ \text{then:}$$ $$\forall \epsilon, \ (\epsilon > 0) \implies \exists \delta \ (\delta > 0 \ \text{and} \ \forall x, \ ((x\neq a \ \text{and} \ |x-a| < \delta) \implies |f(x)-L| < \epsilon))$$ I have absolutely no idea what this means. I think I get the first part, which is:

"For all $\epsilon$, if $\epsilon > 0$, then there exists a $\delta$ such that $\delta > 0$..."

From that point on I do not get it. I am not sure if what I wrote above is even correct. I am worried that this may be a very important definition to understand and memorize, so I need your help understanding it. A few questions that I have are:

$1$. Where did the $\epsilon$ and the $\delta$ even come from?

$2$. Why does $\epsilon$ and $\delta > 0$?

$3$. What are the absolute value signs for?

I would greatly appreciate some help, thanks!

  • $\begingroup$ Though it doesn't make any real difference, usually people use $\delta$ instead of $\Delta$. $\endgroup$ – mojambo Mar 16 '14 at 23:15
  • $\begingroup$ @D.Clark Understood. Edited the question $\endgroup$ – TrueDefault Mar 16 '14 at 23:17
  • $\begingroup$ FWIW, so did Halmos. $\endgroup$ – user27325 Sep 11 '15 at 17:27
  • $\begingroup$ Absolute value specifies distance. For example, $|x-c| < e$ means $x$ and $c$ are within $e$ of each other (assuming $e>0$) $\endgroup$ – MathematicsStudent1122 Jan 30 '17 at 23:16

Seth's answer here does more than enough to expose the intuition behind the definition of a limit. But I get the feeling you are having trouble untangling the definition itself. So I think this could be useful. Here is what it says:

"Forany given $\epsilon \gt 0$, however small, there is (and there always is) a corresponding positive quantity $\delta$ such that the value of the function $f(x)$ is less than $\epsilon$ distant from $L$ whenever the $x$ values are less than $\delta$ distant from $a$".

No matter how close you want your function to be to $L$ there is what we call a "neigbourhood" made of all points whose distance from $a$ is $\delta$ such that the value of the function $f(x)$ is as close as you previously wanted to $L$ for every value of $x$ in the so stipulated $\delta$ - neighbourhood.

If this is true then we say $L$ is the limit of the function $f$ as $x$ tends to (approaches) $a$.

All of this obviously is given in mechanical terms above. You have untangled it to a certain extent CORRECTLY. There onwards,

The $\forall x$ to me is superfluous. What you have inside the brackets is $|x - a| \lt \delta \implies |f(x) - L| \lt \epsilon$ which essentially means If $|x - a| \lt \delta $ then $|f(x) - L| \lt \epsilon$ or $|f(x) - L| \lt \epsilon \;\;$ whenever $|x - a| \lt \delta$ as I have explained above.

Hope I helped.

  • $\begingroup$ A possibly useful exercise is to assume the contrary and determine what the logical expression means when limx→af(x) =/=L $\endgroup$ – Confutus Mar 17 '14 at 5:37

The first thing to understand is that $|x-y|$ is the distance between $x$ and $y$.

The idea is that epsilon represents an arbitrarily small amount of closeness to $L$, so epsilon is chosen positive because distances are positive. You should think of epsilon being very small. We are saying that for any small distance epsilon from $L$ if we look only at values in the domain of $f$ that are very close (within a distance of delta) to $a$ then their images through $f$ will be very close (within a distance of epsilon) to $L$. The key is that for every value of epsilon we can find some small enough delta. So if epsilon is really small we might have to choose delta really small but we can still find one.

So the point is that if we want $f(x)$ to be very close to $L$ then we can guarentee that this will happen when $x$ is very close to $a$. However, we don't require that $f(a)$ itself is equal (or close) to $L$ because we want to consider only what happens to $f(x)$ as $x$ "approaches" $a$.

This is probably best illustrated with a picture which I of course can't draw here. Perhaps your teacher or someone else could help you with this in person.

I realized while writing this that I could go on and on trying to clarify this concept but I think I'll just stop here and hope that I made things a little clearer. It is important to look at a lot of examples of functions and understand if they are continuous or not. It's worth really thinking this over until it makes intuitive sense to you.


The formal definition of the limit is to say, given an epsilon, I imploy delta that implies, further down the line.


In very simple terms, the delta-epsilon values represent the radii of OPEN intervals on the x and y-axes, which explains the strict inequality. So, we have (c-r, c+r) and its image (L-r, L+r) and you can see why the Greek letters are used. Like a deflating balloon, the interval (c-r,c+r), whose midpoint or center c is r units from either endpoint, approaches its natural limit of zero diameter as the radius r approaches zero. Of course, if any difference a-b = 0, a=b is a single point where zero distance naturally occurs. When the domain interval "deflates like a balloon," the diameter (and thus the radius) approaches its natural limit of zero that occurs at the point c. By this, the image (reflection, AVATAR, shadow etc) has no choice but to approach L as delta approaches zero. By definition, delta is greater than zero so we can avoid embarrassment in the case of a function such as y = 1/x and c =0.

To see the connection, let x be a point in (c-r, c+r), then x is between c-r and c+r. By subtracting c from both sides of the inequality, we have x-c is between -r and r, which is the definition of the absolute value of x-c.


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