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How do you solve a limit

  1. Feb 28, 2008 #1
    hi im new here so i was wondering if anyone can tell me instructions on solving a limit any limit
  2. jcsd
  3. Feb 28, 2008 #2
    First, welcome to PF!
    Well, there are different types of limits, and i think the best way you could grasp an idea of what limits are is to find a book that has some kind of introduction to limits, and start reading it. So if you have any particular question about limits, that is any specific question on how to evaluate a specific limit it would be easier for us to help u, cuz limits are a broad topic. However, here is an example: the limit of any polynomial as x-->a, as x goes to a number a, is merely the value of that polynomial at that point:
    [tex]lim_{x\rightarrow \ 2}(x^{3}+4x^{2}-3x)=(2^{3})+4 (2^{2})-3(2)[/tex]
    But there are many,many other types!
  4. Feb 29, 2008 #3


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    Unfortuately, many beginning students, seeing examples like that, get the impression that "limit" is just a fancy way of talking about the value of the function! That is completely untrue- in fact, in general the limit of function, at, say, x= a, has no relationship to the value there.

    The confusion comes because functions for which the limit is the value of the function, called "continuous" functions, are so nice and simple, that we tend to use them all the time! In fact, "almost all" functions are not continuous but almost all the functions we use are continuous.

    janeba, this is a very large question (much like your previous question about derivatives!). The "limit" concept really talks about what happens close to a point but there are a few basic properties you can use:
    1) If f is a constant function, f(x)= C, then [itex]\lim_{x\rightarrow a}[/itex]f(x)= C for all a.
    2) If f(x)= x, then [itex]\lim_{x\rightarrow a}[/itex] f(x)= a.
    Those are called the "trivial limits". Notice that if f(x)= C, then f(a)= C and if f(x)= x, then f(a)= a so those functions are "continuous" at all a.

    3) If [itex]\lim_{x\rightarrow a}[/itex]f(x)= L and [itex]\lim_{x\rightarrow a}[/itex]g(x)= K then [itex]\lim_{x\rightarrow a}[/itex](f(x)+ g(x))= L+ K.
    4) If [itex]\lim_{x\rightarrow a}[/itex]f(x)= L and [itex]\lim_{x\rightarrow a}[/itex]g(x)= K then [itex]\lim_{x\rightarrow a}[/itex](f(x)*g(x))= L* K.
    5) If [itex]\lim_{x\rightarrow a}[/itex]f(x)= L and [itex]\lim_{x\rightarrow a}[/itex]g(x)= K and K is not 0, then [itex]\lim_{x\rightarrow a}[/itex](f(x)/g(x))= L/K.

    Notice that the last does not help you if the limit in the denominator is 0! Yet, the formula for the derivative always involves a fraction with the denominator going to 0! The last property, and I think the most important, although it is seldom emphasized (I remember a text titled "Calculus for Economics and Business Administration" that did not even mention it!) is

    6) If f(x)= g(x) for all x except x= a, then [itex]\lim_{x\rightarrow a}f(x)= \lim_{x\rightarrow a} g(x)[/itex].

    You should be able to see how the first 5 of those give the result for polynomials that sutupidmath mentioned- that all polynomials are continuous.

    But suppose we were given f(x)= [itex](x^2- 4)/(x-2)[/itex], f(2)= 100, and asked to find the limit as x goes to 2. (Notice that [itex](2^2- 4)/(2- 2)= 0/0[/itex] which is "indeterminant". That's why I gave the separate definition of f(2).) That's a fraction so we would want to use (5) above but we can't. The limit in the denominator is just (by (1) above) 2- 2= 0. What we can do is say that [itex]x^2- 4= (x- 2)(x+ 2)[/itex] and as long as x is not 2[/itex] we can cancel the (x-2) in the numerator with the (x-2) in the denominator as say [itex](x^2-4)/(x-2)= (x-2)(x+2)/(x-2)= (x+2)[/itex] as long as x is not 2.

    When x= 2, f(x)= 100 which is certainly not equal to 2+ 2= 4. But that doesn't matter, since f(x)= x+ 2 for all x except 2, [itex]\lim_{x\rightarrow 2} f(x)= lim_{x\rightarrow 2} x+ 2[/itex] and the limit on the right is just 2+ 2= 4 so [itex]lim_{x\rightarrow 2} f(x)= 4[/itex] even though f(2)= 100.

    Of course, if I were to take x= 1.9999 or x= 2.00001, since those are not equal to 2 I would use the first formula and I would get [itex](1.9999^2- 4)/(1.9999- 2)= -.00039999/-.0001= 3.9999[/itex] and [itex](2.00001^2- 4)/(2.00001- 2)= 0.0000400001/.00001= 4.00001, both very close to 4. The limit tells me what a function is like close to but not at a particular value of x.

    (Thanks for not posting in all capitals!)
  5. Feb 29, 2008 #4
    If you have a continuous differentiable function, you can simply evaluate it at that point, however there are other examples, such as [itex]\lim_{x\rightarrow 1}(x^2-1)/(x-1)[/itex] which can be solved by factorizing the top and canceling the denominator, which gives you two. There are also limit to infinity of rational functions, eg [itex]\lim_{x\rightarrow \infty} (x^2-2)/(2x^2+x-1)[/itex] which is solved by mutiplying it by [itex](1/x^2)/(1/x^2)[/itex] , which gives you [itex]\lim_{x\rightarrow \infty} (1-2/x^2)/(2+1/x-1/x^2) = 1/2[/itex]. There is of course the rigorous epsilon-delta definition, but that is for another day.
  6. Feb 29, 2008 #5
    thank you but i have to learn something before limits
  7. Mar 1, 2008 #6


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    Yes, you have to learn algebra!

    (But I don't think that's what you really meant to say!)
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