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Simple limit question

  1. Nov 14, 2007 #1
    Hi all. I encountered the following problem:

    lim x->0 x * cot(x). My calculator gives the value of 1 for this limit, which can be corroborated by looking at the graph of y = x * cot(x).

    However, I reason that it should be zero, because:

    lim x->0 x * cot(x) = (lim x->0 x) * (lim x->0 cot(x))
    = 0 * lim x->0 cot(x)
    = 0 (because anything multiplied by zero is zero.)

    Clearly, my reasoning is flawed. What is wrong with it? And what is the correct way to do this problem?

    Thank you.
     
  2. jcsd
  3. Nov 14, 2007 #2

    Gib Z

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    Homework Helper

    In general, [tex]\lim_{x\to a} f(a)g(a) \neq (\lim_{x\to a} f(a)) (\lim_{x\to a} g(a))[/tex]

    For example, as x approaches zero, (x/x) approaches 1. If you split it like you did, we get zero times infinity. Clearly "anything multiplied by zero is zero." does not hold.
     
  4. Nov 15, 2007 #3
    it's as Gib Z says, but just so you are clear, lim fg = limf limg IF limf and limg both exist, and the same goes for sums, lim(f + g) = limf + limg if both limf and limg exist, this is a common mistake. so think about this so you don't make it again
     
    Last edited: Nov 15, 2007
  5. Nov 15, 2007 #4

    HallsofIvy

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    That statement is wrong: 0 multiplied by [itex]\infty[/itex] is not 0! It is correct that 0 multiplied by any real number is 0 but that limit is not a real number. The others said that "lim x->0 cot(x)" does not exist which is better wording for the same thing.

    cot(x)= cos(x)/sin(x). x cot(x)= x cos(x)/sin(x)= (x/sin(x))(cos(x)). It is proved in Calculus that [itex]\lim_{x\rightarrow 0} sin(x)/x= lim_{x\rightarrow 0}x/sin(x)= 1[/itex]. Since cos(x) is continuous and cos(0)= 1, the limit of x cot(x)= 1.
     
  6. Nov 15, 2007 #5
    You can use L'Hospital's rule for x/tan(x) goes to 1/(secx)^2 upon derivation top and bottom, which goes to 1 as x goes to 0.
     
    Last edited: Nov 15, 2007
  7. Nov 15, 2007 #6
    Thanks a million, both of you!
     
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