I am in a total delusion about understanding limits relationg to real world elements. I am going the lecture notes again and again. So here Just not only I don't get delta epsilon approach it confuses me more.

  1. What is a limit?
  2. On what basis it is called a limit? Is there a better word to describe it? Is it called limit because it is limiting something that has impact on another? Such as $f (x)$ is a function of $x$. Now $x$ has a domain it could move around to make sure that $f (x)$ will exist? If $x$ doesn't comply with the domain then $f (x)$ will not exist? That means $f (x)$ is the limit and $x$ makes it to exist or not. Is that correct?

All I understand is that $x$ has forbidden values which shouldnt be plugged into certain functions of $x$. But since we are curious we want to know the last $x$ value before x hits the forbidden city. If that last point can make function of x to exist without going to infinity then can I say at this last point, the function exists? Why does it call limit I still don't get it. Is it because it is limiting x from making a mess in function of x?

  1. Why do we need to find limits?

  2. When obviously $x=-2$, function of $x$ $1/(x+2)$ is not going to be "okay". It will go to infinity, so can it be marked as not defined and $f(x)$ doesn't exist at $x=-2$? Should we just leave it like that or do we need to find "aid" that will not result in infinity?

Please have some mercy as I am really struggling to get the concept of limits right and stable.

Already checked on two questions here

  • $\begingroup$ See if these Khan Academy videos help. Disclaimer: I haven't listened to them. $\endgroup$ – Dave L. Renfro Sep 19 '13 at 19:54
  • $\begingroup$ I want to vote up everyone who helped me out here. But Guess what I need 15 min rep. So wait will ya. I will vote up and accept the answer that I am able to digest at this level. :) $\endgroup$ – aspiring Oct 24 '13 at 14:44
  1. Consider the sequence:

$$ 1,0.1,0.001,0.0001,0.00001,0.000001,\dots $$

I hope you agree with me that, in some intuitive way, this sequence goes to $0$. It never reaches $0$, of course, but the terms get closer and closer to $0$, without reaching it. In mathematics, we say that the limit of this sequence is $0$.

Notice that it is not necessary for the terms to get closer to $0$ all the time. For instance, we might say that the following sequence goes to $0$ as well:

$$ 1,2,4,1000000,500,100,5,10,0.5,1,0.05,0.1,0.005,0.01,0.0005,0.001,\dots $$

This sequence gets absolutely huge at the beginning, but then starts to get smaller. It doesn't get smaller every step - in fact, the number doubles every other step - but we still have the intuition that it's getting closer to $0$ all the time.

How can we formulate a precise definition for when a sequence $(a_n)$ goes to $0$? One obvious try is to say that $a_\infty=0$. But that isn't a help at all. What is $a_\infty$ anyway? Why can't we just define $a_\infty$ to be $1000$ in the first sequence?

So we're going to have to look at the terms of the sequence with finite index. One thing we might try is: the sequence gets closer and closer to $0$. But, as we have seen above, that doesn't fully capture our intuition of a sequence going to $0$. Besides, the sequence:

$$ 2, 1.1, 1.01, 1.001, 1.0001, 1.00001, \dots $$

gets closer and closer to $0$, but we wouldn't say that its limit is $0$. In fact, it doesn't ever get less than $1$!

It's clearly a problem if the terms of our sequence are never less than $1$, but it's just as much of a problem if the terms of the sequence are never less than $1/1000$, or $1/1000000$, or any other small number you like. In fact, for any small number we like, we need the terms of the sequence to be less (in magnitude) than that number eventually. If we call our arbitrary small number $\varepsilon$, we might end up with the following definition:

Fake definition: A sequence $a_n$ goes to $0$ if for all $\varepsilon>0$, there exists $n$ such that $|a_n|<\varepsilon$.

But that poses problems as well. For instance, we wouldn't say that the following sequence goes to $0$:

$$ 1,0.1,1,0.01,1,0.001,1,0.0001,1,0.00001,1,\dots $$

even though it satisfies our definition above. Instead, we want the terms of the sequence, once they have become smaller than $\varepsilon$, to stay smaller than $\varepsilon$. So we end up with the usual definition:

Definition: A sequence $(a_n)$ goes to $0$ or converges to $0$ or is a null sequence if for every $\varepsilon>0$ there exists $N(\varepsilon)$ such that for all $n\ge N(\varepsilon)$, $|a_n|<\varepsilon$.

We can then give the following definition which is, I hope, fairly clear:

Definition: A sequence $(a_n)$ has limit $a$ (written $a_n\to a$), if the sequence given by $(a_n - a)$ is a null sequence.

  • $\begingroup$ Appreciate everyone's reply. Let me go through carefully. $\endgroup$ – aspiring Sep 19 '13 at 23:55
  • $\begingroup$ Thanks for answering. I waited till I get the ability to vote up. Sorry for the delay. $\endgroup$ – aspiring Dec 3 '13 at 15:29
  • $\begingroup$ No problem! I'm glad I was able to help. $\endgroup$ – John Gowers Dec 3 '13 at 19:26
  • $\begingroup$ Do you think I posted into the correct site for this question? math.stackexchange.com/questions/590786/… I haven't gotten any comments or replies yet... $\endgroup$ – aspiring Dec 4 '13 at 1:52
  • $\begingroup$ @aspiring No - that's a programming question, and you should post it on Stack Overflow instead. $\endgroup$ – John Gowers Dec 4 '13 at 7:00

When a mathematician says the "limit of $f(x)$ is $L$ (as $x$ approaches some number $c$)", he/she means something quite complex. He/she means that when some $x$-values that are close to $c$ (but not actually $x=c$) are plugged into $f(x)$, then the values of $f(x)$ are close to $L$. For example, we feel confident that $\displaystyle{\lim_{x \to 3} x^2 = 9}$ because $(2.99)^2 = 8.9401$, and $(2.9999)^2 = 8.9994$, and $(3.0000001)^2 = 9.0000006$. In each of these calculations, I plug in a number close to $x=3$ and expect the result to be close to $y=9$. Roughly speaking, the word "limit" refers to the ultimate end result of the process of plugging in numbers $x$ that are closer and closer to the target.

But why can't you just plug in $x=3$ into $x^2$ to find out $(3)^2 = 9$, I bet you're wondering. Yes, in this case we can, because $y=x^2$ is nice and continuous, but not all functions are quite so nice. We may also want to examine what are the values of $f(x) = \frac{\sin x}{x}$ near $x=0$. In that case, plugging in $x=0$ does no good, so we go the long way around; plug in numbers that are close to $x=0$ (but not equal to $0$).

$$ \begin{align*} x = 0.1 &\Rightarrow \frac{\sin 0.1}{0.1} \approx 0.9983341665\\ x = 0.01 &\Rightarrow \frac{\sin 0.01}{0.01} \approx 0.9999833334\\ x = 0.001 &\Rightarrow \frac{\sin 0.001}{0.001} \approx 0.9999998333\\ \vdots & \end{align*} $$

After some consideration and calculations, we feel confident that there is some end result, some limiting value to which these numbers approach, a limit to be brief, and that value seems to be $y=1$. (ok, technically, this just gives us evidence that the limit (as $x=0$ is approached from the right) is probably very close to $1$).

The $\epsilon$-$\delta$ approach is a way to make the process rigorous. What does it really mean to approach a value closer and closer How close is close enough? I would encourage you to get a good conceptual grasp on how limits work before tackling the $\epsilon$'s and $\delta$'s.

Hope this Helps!

  • $\begingroup$ Appreciate everyone's reply. Yes i want to and I am digging on that. Can you please give me a real world example (simple) that shows the existence of limits and why checking limits are required? Or make us check them? How to identify limits? $\endgroup$ – aspiring Sep 19 '13 at 23:58
  • $\begingroup$ Thanks for answering. I waited till I get the ability to vote up. Sorry for the delay. $\endgroup$ – aspiring Dec 3 '13 at 15:30

The limit of a function to a particular value describes the behaviour of the function as the function value approaches that value. As an example the following

$\displaystyle \lim_{x\rightarrow x_{0}} f(x) = c \tag{1}$

implies that as $x$ approaches the value $x_{0}$ that the function $f(x)$ takes the value $c$. To demonstrate this we could say $f_{1}(x) = x + 2$, $x_{0} = 2$ and $c=4$ when $f_{1}(x)$ is a function on the real numbers. There are also situations when the limit does not exist. For example consider the function

$f_{2}(x) = \displaystyle \begin{cases} 1, \quad x \geq 0 \\ 0, \quad x < 0.\end{cases}$

Now how do we define the limit of this as $x\rightarrow 0$ i.e., what is $\lim_{x\rightarrow 0} f_{2}(x)$? The answer is that it doesn't exist, as depending on whether we approach $0$ from above or below we get different answers. To show this we introduce the notation $0_{\pm}$ to mean either the 'right' side of zero ($0_{+}$) or the 'left' side of zero ($0_{-}$).

Now we see that if we approach $x=0$ from the right, i.e.

$\displaystyle \lim_{x\downarrow 0_{+}} f_{2}(x) = 1$

but if we approach it from the left, i.e.

$\displaystyle \lim_{x\uparrow 0_{-}} f_{2}(x) = 0.$

The left hand limit does not equal the right hand limit, so the limit at zero is not well defined for $f_{2}$. This characterizations tells us that $f_{2}$ is a discontinuous function. In contrast $f_{1}(x) = x + 2$ is a continuous function as its left limit equals its right limit

  • $\begingroup$ Appreciate everyone's reply. Let me go through carefully. $\endgroup$ – aspiring Sep 19 '13 at 23:54
  • $\begingroup$ Thanks for answering. I waited till I get the ability to vote up. Sorry for the delay. $\endgroup$ – aspiring Dec 3 '13 at 15:28

Limits start off with Pythagoras. Take a unit square. Draw the diagonal. How long is it? We can give it a name, it behaves like we expect, but does it exist and make sense.

If we only know rational numbers, we can get close, but we can't get equal.

Can we get systematically close - closer at each stage than the last one. Turns out that's not difficult.

What if we go on for ever ... ?

This is a real length. Surely we can measure it with a real number?

Conflict resolution: limits exist.

  • $\begingroup$ I guess I need more experience to appreciate the type of explanation you have given. When that day comes, I will smile and think further :D hopefully can reply you with a better comment, a side question. $\endgroup$ – aspiring Oct 24 '13 at 14:43
  • $\begingroup$ Thanks for answering. I waited till I get the ability to vote up. Sorry for the delay. $\endgroup$ – aspiring Dec 3 '13 at 15:30

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