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Indeterminate forms

  1. Jun 1, 2015 #1
    Here's a list of all the indeterminate forms I'm familiar with:
    ##\frac{0}{0}##, ##\frac{\infty}{\infty}##, ##0⋅\infty##, ##\infty - \infty##, ##0^0##, ##1^{\infty}##, ##\infty^0##
    Suppose we want to evaluate the limit:
    $$\lim_{x→0} x^2 \cos{\frac{1}{x}}$$
    We can find the value of this limit by applying the squeeze theorem. The limit would otherwise be indeterminate; if we plug in ##x = 0##, we get:
    $$0⋅\cos{\frac{1}{0}}$$
    Under what category does this indeterminate form lie?
     
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  3. Jun 1, 2015 #2

    ShayanJ

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    This is not an indeterminate form. It has a unique limit: 0.
    The reason is, whatever the argument of the cosine is, we know that the result lies between -1 and 1. So we know that we are multiplying zero by a number between -1 and 1 which is obviously zero.
     
  4. Jun 2, 2015 #3
    I'm confused. The limit of, say, ##x^3 - x^2## as ##x## approaches infinity is also unique; it's infinity. However, at first glance, it seems to be indeterminate (##\infty - \infty##).
    I only arrived at the result by factoring out ##x^3##.
    Perhaps the definition of "indeterminate" is not so clear to me.
    Could you please elaborate further?
     
  5. Jun 2, 2015 #4

    ShayanJ

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    The difference is that, in the case of indeterminate forms, at first you encounter one of the seven forms. Then you should find a way to see what's the value of the indeterminate form in this particular example. Its sometimes finite and sometimes infinite and sometimes doesn't exist. About your example, we first encounter ## \infty - \infty ## which is indeterminate i.e. we don't know its value. Then we search for an alternative way to find its value and we figure out that because ## x^3 ## rises faster than ## x^2 ##, at infinity they should be very far apart which means the answer is infinity.
    But about ## \displaystyle \lim_{x\to 0} x^2 \cos{\frac 1 x} ##, we can directly find out that the answer is zero. Because cosine is always between -1 and 1, and here is no different. We just don't know what is the value of the cosine but we know its between -1 and 1. So just call its value a. Then we have ## \displaystyle \lim_{x\to 0} x^2 a ## which is obviously zero.
     
  6. Jun 2, 2015 #5

    ShayanJ

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    This page may be helpful.
     
  7. Jun 3, 2015 #6

    HallsofIvy

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    This particular function has limit, as x goes to infinity of 0, but there are other functions, such as [itex]x^2- (x^2+ 1)[/itex], that are "of the form" [itex]\infty- \infty[/itex] that converge to other limits. "[itex]\infty- \infty[/itex]" and "0/0" are "indeterminate" because you cannot determine such a limit by just setting [itex]x= 0[/itex] or [itex]x= \infty[/itex] in the sequence.
     
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