1. Sep 22, 2012

### vysero

Okay so here is the deal, I understand the concept behind a limit just fine. However (for some reason) whenever I am asked to graph or visualize the concepts of these notations that my book is using I am failing. So that makes me think that maybe I do not understand limits... for instance my book says:

"To define limits, let us recall that the distance between two numbers a and b is the absolute value of a - b, so we can express the idea that f(x) is close to L by saying that the absolute value of f(x) - L is small."

Why does this make no sense to me? What does subtracting a function say: f(x) = x^2-x-2 from L (which i assume (L) means the limit of a function) have to do with a limit of a function. Does it have something to do with the limit of the function that we subtracted the limit of that function from.. I just do not get it!!!!

So, the lim as x approaches c of f(x) = L

what does L stand for... does it stand for the actual number c or does it stand for some abstract concept about taking the limit of c? What does c stand for, and for that matter does f(x) still mean the y axis??... Am I suppose to be thinking of two different graphs here???? I am so confused.

I failed my first calc test... this is the first time in my life I have failed a math test... I studied, I always study but they confused the crap out of me there was no actual math it was all this conceptual ways of thinking about the notations of limits....

For instance:

Assume that: the lim as x approaches infinity of f(x) = L and the lim as x approaches L of g(x) = infinity,

true/false:

A) x = L is a vertical asymptote of g(x).
B) y = L is a horizontal asymptote of g(x).
C) x = L is a vertical asymptote of f(x).
D) y = L is a horizontal asymptote of f(x).

My book mentioned one thing about asymptote's and that is this rule: "A horizontal line y = L is a horizontal asymptote if: the lim as x goes to infinity of f(x) = L and/or the lim as x goes to negative infinity of f(x) = L"

Guess which one of those questions i got right... is this some new thing about math in calculus I now have to be able to manipulate the books words and notations in order to figure this stuff out all on my own....? I mean didn't it takes thousands of years before one or two guys figured this out on there own?????? They expect us all to be as good at math as Newton or am I just completely missing something about limits and the manipulation of theorems. Maybe I just have no imagination and so I can go no further in math....???

I know this is a lot and I don't expect someone to answer all of my questions but I will take any help on translating this notional crap that's in my book into something useful that I can actually work with and understand.

2. Sep 22, 2012

### jbunniii

x and c are numbers on the x axis.

f(x) and L are numbers on the y axis.

To say that $\lim_{x \rightarrow c} f(x) = L$ means that if $x$ is sufficiently close to (but not equal to) $c$, then $f(x)$ will be close to $L$.

3. Sep 22, 2012

### LCKurtz

OK, I'll give it a shot. Remember that a function is a mapping from a domain, say the $X$ axis, to a range, say the $Y$ axis. You would call the mapping f. For each $x \in X$ you get a number $y\in Y$. We usually write $y = f(x)$ for this, meaning that for a given $x$ on the X axis, the function $f$ maps it to $y = f(x)$ on the $Y$ axis. So if you pick a particular number $x$, $f(x)$ is just a number. To say that it is close to $L$ just means that if you mark $f(x)$ and $L$ on the $Y$ axis, they are close to each other so $|y -f(x)|$ is small.
Now what the statement$$\lim_{x\rightarrow c}f(x) = L$$means is that if you take the variable $x$ and move it closer and closer to $x= c$ on the $X$ axis, the number $f(x)$ will vary accordingly and get closer and closer to $L$ on the $Y$ axis. Here's a simple example. Consider the function $y = f(x) = x^2$. Take out your calculator and calculate $f(x)=x^2$ for the numbers 1, 1.5,1.8,1.9,1.99,1.999. These are numbers getting close to $c = 2$. What does $f(x)$ get close to? I will call your answer $L$. That's the idea of$$\lim_{x\rightarrow 2} x^2 = L$$When you are done you will likely see an easier way of calculating the answer for $L$ in this example. Next try the same idea for this limit$$\lim_{x\rightarrow 0}\frac {\sin x}x$$ Pick some number $x$ nearing $0$ and see what this function nears. Remember to have your calculator in radian mode. After you do that graph it on your calculator and see if the answer makes sense.

That's all I have time for right now. Come back after you try that and we'll talk more.

4. Sep 22, 2012

### vysero

Ok so the first answer is 4 and the second answer seems to have something to do with 1 or maybe it is 1.

lim of f(x) as x approaches infinity = L, if absolute value of f(x) - L becomes arbitrarily small.

Now in the x^2 example you gave me the only thing I can think of that is becoming arbitrarily small is the difference between the number that x is approaching (which is 2) and my x values that I am inserting. So that is the difference between L and f(x) in this case? Which is 2 is f(x) 2? Still confused obviously, ile keep thinking but maybe you can clear it up real quick.

5. Sep 22, 2012

### LCKurtz

Yes to both. For $x^2$ you may have noticed that what it is getting close to as $x$ gets close to $2$ is in fact the value of the function when $x=2$. Doesn't it make intuitive sense to you that if $x$ is close to $2$ that $x^2$ would be close to $4$? Lots of functions are like that -- the limit of the function as x nears $c$ is the functions value at $c$? You will see shortly in your course that functions that behave like that are called continuous functions.

You are also seeing that$$\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$$ but in that case you can't get the answer by plugging in $x=0$ like you could for $x^2$.
When $x$ gets close to $2$, does't $x^2$ get close to $4$? In this case $|f(x) - L| = |x^2-4|$.

6. Sep 23, 2012

### vysero

Yes when you put it that way it makes absolute sense. Even though x^2 can = any positive number since we are limiting it to 2 its kind of like those x values that were coming out in fact it is those values.

I guess what is confusing me is I am use to the negative sign meaning literally subtracting values not just meaning "gets close too". Cause if I were to subtract 4 from one of those values I was plugging into f(x) or x^2 in this case I would just get like well for instance 1.99-4=-2.01 and abs of -2.01 is just 2.01.

7. Sep 23, 2012

### LCKurtz

No. You would get $|1.99^2-4|$. It's $f(x)$ that goes there, not $x$.

8. Sep 23, 2012

### vysero

Oh I see now lol that makes sense. So I think I understand what this means now:

lim of f(x) as x approaches infinity = L, if absolute value of f(x) - L becomes arbitrarily small

thank you!

9. Sep 23, 2012

### LCKurtz

We have been talking about $x\rightarrow c$, not $\infty$. But, yes, $|f(x)-L|$ becomes small as $x$ nears $c$. If you are talking about limits as $x\rightarrow \infty$, then $|f(x)-L|$ gets small as $x$ gets large, which isn't the same situation as $x$ nearing a finite number $c$.

10. Sep 23, 2012