Okay, I get the basic idea of a limit: the value that the function "wants" to reach, as x (or whatever the independent value is designated) approaches the "arrow number" (whatever it's actually called). And when I deal with many limit problems, my answers happily agree with the text's answers; these are the "no duh" type problems, where you just substitute a value in for x, and the limit answer pops out. No problem. But then there are other times, when the answer would seem to be obvious, but my answer is totally wrong. This means that I must be missing something pretty basic/fundamental about limits, but I don't know what it is. Here's an example of what I'm talking about: In Mark Ryan's book, Calculus Workbook for Dummies, on p.43, he present the following problem: limit (as x approaches 0) of x/(sin 3x) = ???. The answer, shown on p.52, is 1/3. He then proceeds to shows how he arrived at that answer, which all seems logical. Logical, that is, as far as following his step-by-step solution, but just not logical, overall...at least to me. So here's my question: how can it be that the answer to that limit problem is 1/3, when, if you simply substitute 0 for x, you would get 0/0, which would indicate that the limit does not exist? So, I guess what I am asking, in a more general sense, is: why doesn't substitution always work, when doing limits? Obviously, it must have to do with the idea that the "x/arrow number" (what the **** is that called, anyway?) is approaching a given value, such as 0, but not actually equal to it. If anyone could shed some light on what would seem to be a pretty basic idea I would really appreciate it, as I hate the idea of just learning by blindly following some "cookbook," step-by-step plan/algorithm, without really knowing what/why I'm doing it, because that's not really learning. Thanks, in advance, for your help. I'm trying to figure calculus out by myself, no class, no teacher, and I'm not exactly a Karl Friedrich Gauss, to start with.
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I'm Just Not Getting Something Really Basic About Limits
- Thread starter rayroshi
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Bob Brown MSEE
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problem: limit (as x approaches 0) of x/(sin 3x) = ???.
So, I guess what I am asking, in a more general sense, is: why doesn't substitution always work, when doing limits? Obviously, it must have to do with the idea that the "x/arrow number" .
Lets call it f(x) = x/(sin 3x)
Lim f(x) as x->0 can exist because we never get there.
EVEN THOUGH f(0)=0
Look at this example g(x) = (x+1)(x-1)/(x-1)
Here it is clear
Lim f(x) as x->1 is 2
(you would like to cancel the (x-1) but you can't when x=1)
It helps if you get a good graphing program. It is useful in Calculus to see the function.
My example is a simple straight line with a hole at (x,y) = (1,2)
Last edited:
OK Here is an attempt.
Let's say h(x) is continuous at a. Now that is DEFINED to mean
\(\displaystyle \displaystyle h(a) \in \mathbb R,\ and\ \lim_{x \rightarrow a^+}h(x) = \lim_{x \rightarrow a^-}h(x) = h(a).\)
So when you are dealing with a function that is continuous at a point, you can find the limit of the function at that point simply by calculating the function at that point.
What happens when you are dealing with a function, say g(x) that is not continuous at some point, say a, because g(x) is not defined at a?
In that case the limit may or may not exist. No problem if the limit does not exist. But it does seem sort of weird that the limit of the function may exist at a point where the function itself does not exist.
The explanation is that, strictly speaking, the limit of g(x) at a always involves values of x that are close to but not equal to a. That is why it is said the limit as x approaches a. But that also explains why the limit can exists as x approaches a even though the function does not exist at a. We specifically do not let x = a.
So what do we mean by the limit of g(x) as x approaches a. If such a limit exists, it is the number that g(x) is very close to when x is very close to a. Now obviously, if g(a) does not exist, then you can't find the limit of g(x) as x approaches a by plugging in g(a). But g(x) may be quite well behaved when x is close to a but not equal a.
Go back and look at the delta epsilon formulation of limit with this hopefully intuitive, non-rigorous explanation as a guide. Feel free to ask further questions.
Let's say h(x) is continuous at a. Now that is DEFINED to mean
\(\displaystyle \displaystyle h(a) \in \mathbb R,\ and\ \lim_{x \rightarrow a^+}h(x) = \lim_{x \rightarrow a^-}h(x) = h(a).\)
So when you are dealing with a function that is continuous at a point, you can find the limit of the function at that point simply by calculating the function at that point.
What happens when you are dealing with a function, say g(x) that is not continuous at some point, say a, because g(x) is not defined at a?
In that case the limit may or may not exist. No problem if the limit does not exist. But it does seem sort of weird that the limit of the function may exist at a point where the function itself does not exist.
The explanation is that, strictly speaking, the limit of g(x) at a always involves values of x that are close to but not equal to a. That is why it is said the limit as x approaches a. But that also explains why the limit can exists as x approaches a even though the function does not exist at a. We specifically do not let x = a.
So what do we mean by the limit of g(x) as x approaches a. If such a limit exists, it is the number that g(x) is very close to when x is very close to a. Now obviously, if g(a) does not exist, then you can't find the limit of g(x) as x approaches a by plugging in g(a). But g(x) may be quite well behaved when x is close to a but not equal a.
Go back and look at the delta epsilon formulation of limit with this hopefully intuitive, non-rigorous explanation as a guide. Feel free to ask further questions.
D
Deleted member 4993
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Okay, I get the basic idea of a limit: the value that the function "wants" to reach, as x (or whatever the independent value is designated) approaches the "arrow number" (whatever it's actually called). And when I deal with many limit problems, my answers happily agree with the text's answers; these are the "no duh" type problems, where you just substitute a value in for x, and the limit answer pops out. No problem. But then there are other times, when the answer would seem to be obvious, but my answer is totally wrong. This means that I must be missing something pretty basic/fundamental about limits, but I don't know what it is. Here's an example of what I'm talking about: In Mark Ryan's book, Calculus Workbook for Dummies, on p.43, he present the following problem: limit (as x approaches 0) of x/(sin 3x) = ???. The answer, shown on p.52, is 1/3. He then proceeds to shows how he arrived at that answer, which all seems logical. Logical, that is, as far as following his step-by-step solution, but just not logical, overall...at least to me. So here's my question: how can it be that the answer to that limit problem is 1/3, when, if you simply substitute 0 for x, you would get 0/0, which would indicate that the limit does not exist? So, I guess what I am asking, in a more general sense, is: why doesn't substitution always work, when doing limits? Obviously, it must have to do with the idea that the "x/arrow number" (what the **** is that called, anyway?) is approaching a given value, such as 0, but not actually equal to it. If anyone could shed some light on what would seem to be a pretty basic idea I would really appreciate it, as I hate the idea of just learning by blindly following some "cookbook," step-by-step plan/algorithm, without really knowing what/why I'm doing it, because that's not really learning. Thanks, in advance, for your help. I'm trying to figure calculus out by myself, no class, no teacher, and I'm not exactly a Karl Friedrich Gauss, to start with.
The problem comes from the fact that division by 0 is not defined.
If you plot the function y = x/sin(3x), it will become clear to you that the value of the function approaches 1/3 as x approaches 0.
However, self-teaching calculus, without a face-to-face guide is not advisable. Even Gauss had teachers that guided him.
HallsofIvy
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- Jan 27, 2012
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For x not 0, we can write \(\displaystyle f(x)= \frac{1}{3}\frac{3x}{sin(3x)}= \frac{1}{3}\frac{u}{sin(u)}\). where u= 3x.
Now, what do you know about \(\displaystyle \lim_{u\to 0}\frac{sin(u)}{u}\)?
Now, what do you know about \(\displaystyle \lim_{u\to 0}\frac{sin(u)}{u}\)?
I learned limits as a CHALLENGE.
If I say \(\displaystyle \displaystyle \lim_{x \to 0} \dfrac{x}{\sin{3x}} = \dfrac{1}{3}\),
What I mean is "You give me an \(\displaystyle \epsilon\), no matter how small,
.......................and I can give you a \(\displaystyle \delta\) such that if x is within \(\displaystyle \delta\) of 0,
.......................then \(\displaystyle \dfrac{x}{\sin{3x}}\) will be within \(\displaystyle \epsilon\) of 1/3."
If other words, however close a tolerance YOU set on the value 1/3, I can beat it by making x close enough to 0.
If I say \(\displaystyle \displaystyle \lim_{x \to 0} \dfrac{x}{\sin{3x}} = \dfrac{1}{3}\),
What I mean is "You give me an \(\displaystyle \epsilon\), no matter how small,
.......................and I can give you a \(\displaystyle \delta\) such that if x is within \(\displaystyle \delta\) of 0,
.......................then \(\displaystyle \dfrac{x}{\sin{3x}}\) will be within \(\displaystyle \epsilon\) of 1/3."
If other words, however close a tolerance YOU set on the value 1/3, I can beat it by making x close enough to 0.