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How solve limits question.

  1. Jul 2, 2011 #1
    Consider this very simple example of limits
    limitx->2 (x2-4)/x-2
    value of function at x=2 is not defined. but rearranging function like this limitx->2 x+2 gives us it's limit at x=2.
    My question is how rearranging a function gives limit of that function at the specified point..
    I have studied limits fully (as much as it is in my course book) don't go to question heading.
    thanks for answering.
  2. jcsd
  3. Jul 2, 2011 #2
    Let's take your function as an example. We know that
    if [itex]x\not=2[/itex]. Since we are taking the limit as x approaches 2, x does not equal to 2 (it's just very very close), so we can simplify the function. Hence the original limit is equal to the limit of x+2 as x approaches 2 since x+2 is equal to the original function if and only if x isn't 2, which we know is true since we are just approaching 2 without actually reaching it. When x is very very close to 2, x+2 is very close to 4 so the limit evaluates to 4.
  4. Jul 2, 2011 #3


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    One of the properties of limits that isn't always given as much attention as it deserves:

    If f(x)= g(x) for all x except a, then [itex]\lim_{x\to a} f(x)= \lim_{x\to a} g(x)[/itex].
  5. Jul 2, 2011 #4
    what i understand from your answer.

    when we take limit of a function at specified point then we search a function that is similar to that of original but also continuous at the point where we want to get limit. As i say in my last example in that case we search for x+2 this function's behavior is similar to (x^2-4)/(x-2) but it is continuous at x = 2. so by putting x=2 in this function we get our limit.

    Am i correct. make correction if i am wrong.
    thanks for helping.
  6. Jul 2, 2011 #5
    That works in this particular case, but it doesn't work in general.

    What is important about a limit is that we are interested in the behavior in a neighborhood close by to the point of interest, but not necessarily at the point itself.

    So when we write [itex]L = \lim_{x \rightarrow a}f(x)[/itex] we are interested in the behavior of f near a, but not necessarily at a.

    [TeX question. Why does my x->a appear to the right of the lim? How do I put it underneath?]

    In the case of a simple rational expression like [itex]\frac{x^2-4}{x-2}[/itex] we can just cancel [itex]x-2[/itex] to get a familiar polynomial function that we can just evaluate at [itex]a[/itex]. So this example is a little misleading.

    In the general case, you can't always do that. For example assuming that you're in calculus class, you'll soon be shown how to evaluate [itex]\lim_{x \rightarrow 0}\frac{sin(x)}{x}[/itex].

    For that, you need to understand that you are NOT trying to find some other function that you can plug 0 into; but rather, that you are trying to understand the behavior of [itex]\frac{sin(x)}{x}[/itex] NEAR [itex]x = 0[/itex].

    The important concept to understand is that the limit of a function is all about what a function does NEAR a point, but not necessarily AT that point. Keep that in mind as you work through your class.
  7. Jul 2, 2011 #6
    To be more technical, I suggest you use the method known as factorization.

    [itex]lim_{x\rightarrow2}\frac{x^{2}-4}{x-2}=lim_{x\rightarrow2}\frac{(x-2)(x+2)}{x-2}= lim_{x\rightarrow2}x+2[/itex]
    So, now it is very simple, just plug in 2 .

    Limits work because as you approach your x value, in this case 2, the limit sees exactly what y value you are approaching. In truth, there is a hole in the function at x=2. by using the limit, we can calculate the real value at x=2. There is another method known as conjugate multiplication, which mathematicians used to solve the value of e. It is done by multiplying the function by it's conjugate, and then taking the limit.
    So, to solve the limit, you first ignore the limit sign and solve how you would if x was not 2. Then you simply plug in the value stated in the limit. In this case, the number approaches 4, and eventually becomes infinitely close, and when it is infinitely close, the number is in fact 4. That is the wonder of limits and continuity.
  8. Jul 2, 2011 #7
    I apologize for being my usual pedantic self ... but it is totally wrong to speak of anything being "infinitely close." There is no such thing in standard real analysis or calculus.

    The concept of limit was invented to be able to speak rigorously about limits without using vague, imprecise, and undefined language such as two quantities being "infinitely close."

    I feel the need to point this out, because the original poster is learning about limits and is struggling to understand their meaning. In order to learn the limit concept, it's essential to focus on the mathematical epsilon/delta definition.

    Nothing "eventually becomes infinitely close" to anything else. That entire phrase is false and misleading to a student learning the mathematical concept of a limit.
    Last edited: Jul 3, 2011
  9. Jul 3, 2011 #8
    Because you are using itex tags. Use tex tags (at the cost of in-line setting) to get [tex]\lim_{x\to a}f(x) = L[/tex] and here is some text after, appearing on a new line even though the source is on the same line.
  10. Jul 3, 2011 #9
    Use \displaystyle as in: [itex]\displaystyle \lim_{x\rightarrow a} f(x) = L [/itex], where now it's inline.
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