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Limits

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What does the limit indicate for? Like lim x=pi/2. Does it mean only x value less or equal to pi/2?

 

Also, what does Devrivates mean? Give me examples won't hurt.

 

Thanks

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HUH? Your links are no big help to me. They are not even working! Plus, it will only direct me to a crap dictionary.

 

What does the limit indicate for? Like lim x=pi/2. Does it mean only x value less or equal to pi/2?

 

Also, what does Devrivates mean? Give me examples won't hurt.

 

Thanks

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As far as i understand a limit, for example [math]\lim_{x \rightarrow 2}f(x) = 5[/math] means that if you take sufficiently large values close to 2, f(x) approaches 5. This could be thought of graphically. Imagine a straight line through the origin in the positive direction, as you increase the value of x closer to 2, the value on the y axis becomes closer to 5. There are many websites and tutorials which explain limits which i'm sure would be unearthed by a quick google search.

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Limit... the limit on a function f(x) as x approaches a certain value...

 

take x^2 for example... as x goes to zero, the values of the function get closer and closer to 0 from both sides, so [imath]\lim_{x \to 0}x^2=0[/imath]. Now, you might be saying, well it's just the value of the function at that given point, right? well, almost. But that's more complex.

 

A derivative is quite simply the equation for the slope of the tangent line of a function. (denoted f'(x)). So, if f© and f'© both have values, then f'© is the value of the slope of the tangent line to the equation f(x) at the point x=c.

 

Does that help? (If you don't get it, you'll just have to wait till you take Calculus...)

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HUH? Your links are no big help to me. They are not even working! Plus' date=' it will only direct me to a crap dictionary.

 

What does the limit indicate for? Like lim x=pi/2. Does it mean only x value less or equal to pi/2?

 

Also, what does Devrivates mean? Give me examples won't hurt.

 

Thanks[/quote']

 

wow

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Thanks Dr Finlay and BobbyJoeCool. I understand the limit part now, but not the derivative part. I am taking advanced calculus next semester, but I'm very curious about them. I guess I'll do some googling.

 

:P

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Here is the brink of my new reading on "derivatives". The defintion of derivative is "The derivative tells us how to approximate a graph, near some base point, by a straight line". Mainly it is related with the tangent line on graphs. I need a review of tangent lines!! :embarass:

 

Tangent line is between 2 points on a curve slope, get [math]x^2[/math] for instance. I forgot how to find the single middle point on a tangent by some kind of formula?

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My teacher gave a great image for understadning limits and derivatives.

 

Imagine you're a runner and there is someone taking a picture of you about to run over the finish line. they keep taking pictures as you run up to the finish line, and keep taking pictures up to the very instant you are about the cross the line... but the camera runs out of film when you actually cross the line. If you take a look at the pictures at another time, you can see that you are just about to cross the line. You have no proof that you actually did, because there is no picture of it happening but, in the picture you do have you come so close that you can say, with relative certainty, that it you crossed the line. It is possible, however, that you didn't cross the line, maybe you sponteneously combusted first, or accidently stepped into a wormhole... but because you have no proof you did cross the line.

 

This is a limit. As you come infinately close to running over the finish line, you can assume that you do so.

 

How is this represented on a graph, you ask?

 

Imagine you have a curve, doesn't matter which one. Now, as you move along this curve, there is a hole in the line. But, if you approach the hole, but don't come to it (take pictures of the runner, but not crossing over the finish line) you would never know that there is no hole. So the limit as you approah the whole would be the place where the graph would be if there was no whole.

 

On to derivatives... how are limits related to derivatives? First, you know that in order to measure the slope of a line, you need at least two points on that line. So let's say that you have a curve and you know two points on that curve. You can find the slope (change in the y values divided by the change in the x values) of the secant line of the curve...

limsec.gif

 

 

But what if you only had one point on that curve and you wanted to know the slope of the tangent line (the slope of a straight line that touches just one point on that curve

Secant%20and%20Tangent%20lines.gif

 

What you would need to do, is find another point on the curve that you know, and you can find out the slope of the secant line, then bring the other point closer and closer to the point you are given (sound familiar?)

limsec2.gif

limsec3.gif

 

Basically, you bring the second point infinately close to the first one, or you approach it, as in take the limit. Eventually, your two points are so close, that they may as well be considered one point, and your secant line becomes you tangent line. In other words, you take limit as the second point apporaches infinately close to the first one. In this way you can find out the slope of your tangent line, when given only one point... AKA a derivative.

 

Hope this helped. If it didn't please let me know, and I'll try to clarify further.

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Ecoli's post was extremely thorough here, I wish there was a karma system in place for posts like that. Its a primer!

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thanks cosine... (but please remember, you taught me some of my first introduction stuff into calculus)

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I like the explanation of a limit. One of those golden examples i won't forget, which is useful.

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Which point a or x on the graph would approach one other? How do you know? Because at the "a" point, the slope can be significally different than at the "x" point. At "a" the slope would be negative and at the "x" the slope would be zero or teeny positive. ???

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Which point a or x on the graph would approach one other? How do you know? Because at the "a" point, the slope can be significally different than at the "x" point. At "a" the slope would be negative and at the "x" the slope would be zero or teeny positive. ???

 

Because, its a continous function, so there's a patern.

 

EDIT: (c is the distance between x and a)

 

look at that graph... the point (x,f(x)) is one point on the secant line right? and the other is (x+c,f(x+c)) right? and the derivative is the slope of the tangent line right? so, you move the point (x+c,f(x+c)) closer and closer to (x,f(x)) (in other words you make c smaller and smaller until it's zero)... the whole time, the slope of the secant line will be [imath]\frac{f(x+c)-f(x)}{c}[/imath]... being [imath]\frac{\Delta y}{\Delta x}[/imath], right? (please note the c IS the change in x). So, what happens to the graph as you make c closer to 0? The secant line moves closer to being the tangent line, right? and when c is 0? That's right! It's the Tangent line!!! The problem however, is that now that you have a tangent line, the equation for slope has become [imath]\frac{f(x)-f(x)}{0}[/imath], which is 0/0. But, we get around this, because as long as c does not equal 0, the equation for the slope is defined. so we need to find the limit on the function as c goes to zero....

 

So the [imath]\lim_{c \to 0}\frac{f(x+c)-f(x)}{c}=f'(x)[/imath] where f'(x) is equal to the derivitive, which is the slope of the tangent line. In order to get the answer you need to get the c out of the denominator (which can be difficult), but this is the definition of a derivative, and any proof involving derivatives will involve this somehow...

 

Any better?

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Which point a or x on the graph would approach one other? How do you know? Because at the "a" point, the slope can be significally different than at the "x" point. At "a" the slope would be negative and at the "x" the slope would be zero or teeny positive. ???

 

In this graph x is being moved to a... but it doesn't really make a difference, as long as you can understand it conceptually.

 

If you take a limit as x appraches a, the slope will get greater and c - the distance between the two (or more commonly known as "h"), will decrease. If you take a limit as c approaches 0, its the same as taking a limit when you move point x to point c. Can you see that?

 

If you know how to calculate the slope, then you get the definition of a derivative. The change in the y value (represented by f(x) {the formula for calulating the curve} plus "c" subtracted by that same function) divided by the change in the x value, which is just c. As C gets closer and closer to zero, the slope is equal to the tangent line slope.

 

Understand?

 

Well, I have a dentist apointment now, so I'll explain more later.

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Ok... so let's say you were given the equation [imath] y = x^2 [/imath]. You know the formula for the derivative is just a modified equation for the slope of a straight line, when 2 points are known... which is f(x + c) - f(x)/ c... so just substitute your equation.

 

f(x) would equal [imath] x^2 [/imath] and f(x+c) would equal [imath] (x+c)^2 [/imath]

 

and then take the limit (lim) as c approaches zero

mathematically, this would be represented by

[math]

\lim_{c \to 0}\frac{(x + c)^2 - x^2}{c}=

[/math]

 

you then FOIL the top term to get

[math]

\lim_{c \to 0}\frac{(x^2 +2xc + c^2) - x^2}{c}=

[/math]

 

the [imath]x^2[/imath] cancel out and you have

 

[math]

\lim_{c \to 0}\frac{ 2xc + c^2 }{c}=

[/math]

 

which equals

 

[math]

\lim_{c \to 0}\frac{ c(2x + c) }{c}=

[/math]

 

that c can cancel out with the one from the bottom, leaving you with

 

[math]

\lim_{c \to 0}\ 2x + c=

[/math]

 

now it's time to apply the limit. As the variable c approaches zero, the c term in the problem approaches zero... giving you [imath] 2x + 0 [/imath] or simply 2x. This is the equation for your derivative... the equation for the slope of the tangent line to your curve. (or more acurately g(x) = 2x, when g(x) is equal to the derivative of f(x).

 

did that do anything for you?

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I understand a bit more now. Thanks for trying to put in effort to explain it to me. :)

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no problem... that's what this forum is all about.

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I figured since this is a thread on limits, perhaps a little something should be said about the rigorous definition of limits.

 

[math]lim_{x-->a} f(x) = L[/math]

 

Means that for any [math]e > 0[/math] there exists a [math]d > 0[/math] such that

[math]|f(x) - L| < e[/math]

and

[math]|x - a| < d [/math]

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I'm not too sure about that "and". Personally I think it should be a "whenever", because then that statement is equivalent to:

 

[math]\forall \epsilon > 0 \, \exists \, \delta > 0 \text{ such that } 0 < |x-a| < \delta \Rightarrow |f(x) - L| < \epsilon[/math].

 

However, this is pretty much completely beyond the scope of this thread, so if you don't understand it, don't worry about it :)

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ha holy crap, whats with all the greek letters :-s lol, but yah, derivatives are pretty easy when you get what they mean. When you first learn them you have to use the limit definition to get them (which is a pain), eventually though you learn the shortcut:

 

[math] f(x) = ax^n[/math]

[math] f'(x) = (a*n)x^{n-1}[/math]

 

So say you want the derivative of 2x^3. Times the multiplier (2) by 3, take away one from the power and voila! 6x^2 is your derivative. It'd take half a page of work to do it using the limit definition :P

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ha holy crap' date=' whats with all the greek letters :-s lol, but yah, derivatives are pretty easy when you get what they mean. When you first learn them you have to use the limit definition to get them (which is a pain), eventually though you learn the shortcut:

 

[math'] f(x) = ax^n[/math]

[math] f'(x) = (a*n)x^{n-1}[/math]

 

So say you want the derivative of 2x^3. Times the multiplier (2) by 3, take away one from the power and voila! 6x^2 is your derivative. It'd take half a page of work to do it using the limit definition :P

 

or, more generally, you could say [math]y=av^n[/math](where v is another function or combination of them ex. sinx or x^2+x)[math]\frac{d}{dx}y=nav^{n-1}{\frac{d}{dx}}v[/math]

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more generally and more complicated :P lol
not more complicated....more fun! :D

 

there are a lot of little rules like the one CanadaAotS gave that are just special cases of more general rules.

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