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What is the point of limits?

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data

    We have the limit of x -> -1 OF (2x+2/x=1)

    We plug in -1 into the equation and find that it is = to (0/0) therefore it is undefined.
    We then go into attemting to simplify (2x+2/x=1) and we simplify it to (2/1) so now we know that the limit is undefined at -1 but = 2 at any other point. I think I am understand how to these problems (it seems basic so far) but what does this actually mean? How is it applied to something and what does it mean fundementally?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2012 #2
    Well, the limit is defined at x=-1. The limit is 2. In fact, the limit is defined everywhere and is equal to 2 everywhere. For this problem, it means that this function behaves exactly like the function g(x) = 2 EXCEPT it is undefined at x=-1. But for every other point, the two functions are identical. And, not only that, their graphs are exactly the same, except for the hole at x=-1. So, we can define this new function g(x) = f(x) if x isn't -1 and g(x) = lim_{x to -1}f(x) when x=-1. Then we get a nice continuous function.
     
  4. Oct 8, 2012 #3
    So if we have a function like this in real life we would know our out come with be 2 everytime (with exception if we have negitive 1 as x) could you give an example as to how this would apply to a real life situation?
     
  5. Oct 8, 2012 #4

    SammyS

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    The limit is defined. It's the expression, (2x+2)/(x+1), which is undefined for x = -1.
    I assume you mean the limit (as x → -1) of (2x+2)/(x+1),

    i.e. [itex]\displaystyle \lim_{x\,\to\,-1}\ \frac{2x+2}{x+1}\ \ .[/itex]

    There is one small but very important difference between the graph of [itex]\displaystyle y=\frac{2x+2}{x+1}\ \ [/itex] and the graph of [itex]\ \ y=2\ .[/itex]

    That difference is, the behavior of the graphs at x = -1 .
     
  6. Oct 8, 2012 #5
    Yes, that is the expression, thank you. So the difference between the 2 graphs is that at x= -1 it is undefined. So the point of limits is??? I am very sorry I am not understand such an easy topic.
     
  7. Oct 8, 2012 #6

    SammyS

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    One very important application of limits is the "derivative" of a function. you such a thing.

    The derivative of the function, f(x), at x=a, gives the slope of the line that is tangent to y=f(x) at x=a.

    Are you familiar with derivatives?

    If not this Forum is probably not the place to try to teach
     
  8. Oct 8, 2012 #7
    Only slighty, all I know so far is how to find a derivative using the definition of a derivative (is that a proof???) and the power rule.
     
  9. Oct 9, 2012 #8

    HallsofIvy

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    I hope you have realized that you need to learn more about limits!
    As to why we need limits, one reason goes back to Newton himself.

    One of Newton's achievements was finding that the gravitational force between two objects is inversely proportional to the square of the distance between them- the point being that the gravitational force can be calculated at a given instant. The acceleration of an object proportional to that force so you could use that to calculate the acceleration of the object at that instant. But classically, "acceleration" is the change in speed over a given time interval- and "speed" itself can only be defined (classically) over a time interval.

    So how can we be able to calculate acceleration (or speed) at a specific time if they can only be defined over an interval? The answer is that we can take the limit, as the length of the time interval goes to 0, to define "instantaneous" speed and acceleration.
     
  10. Oct 9, 2012 #9

    jbunniii

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    We can even go back to the ancient Greeks, who 2000+ years ago used limits to find the areas and volumes of various objects by approximating them with polygons with arbitrarily many sides. They called this the method of exhaustion, essentially a precursor of the integral.

    Archimedes was able to use this method to approximate the area of a circle (and hence calculate pi) to several digits. He did this by calculating the areas of two polygons with 96 sides each, one inscribed in the circle and the other circumscribed around the circle. Of course he recognized that in principle one could compute arbitrarily many digits by using polygons with more sides.
     
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