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Undefined Points

  1. Jun 19, 2009 #1

    Mentallic

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    I am aware that for a function that is undefined at a point x=a such as [itex]f(x)=1/(x-a)[/itex]

    [tex]\underbrace{lim}_{x\rightarrow a}f(x)=\pm \infty[/tex]

    But it tends to infinite only because it is in the form a/0, where a[itex]\neq[/itex]0.

    Undefined values in the form 0/0 can have a range of values - all reals if I'm not mistaken.

    I thus set up a function f(x) multiplied by another function g(x) so that f(a)=0 and g(a) undefined. However, the functions are not in a form where they can seemingly cancel factors of the zero and undefined value.

    e.g.
    [tex]h(x)=\frac{x+1}{x^2-1}=\frac{1}{x-1}, x\neq \pm 1[/tex]


    So, such a function I simply came up with was

    [tex]h(x)=x*tan(x+\frac{\pi}{2})[/tex]

    I used a graphing calculator to try understand what was happening around x=0, and it seems that

    [tex]\underbrace{lim}_{x\rightarrow 0}h(x)=-1[/tex]

    Now I just want to understand why this limit tends to -1, not any other real values.
     
    Last edited: Jun 19, 2009
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  3. Jun 19, 2009 #2

    arildno

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    Well, you might try utilizing the identity:
    [tex]tan(x+y)=\frac{\sin(x)\cos(y)+\cos(x)\sin(y)}{\cos(x)\cos(y)-\sin(x)\sin(y)}[/tex]
     
  4. Jun 19, 2009 #3

    statdad

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    A simpler example might be

    [tex]
    f(x) = x \sin\left(\frac 1 x \right)
    [/tex]

    for which

    [tex]
    \lim_{x \to 0^+} f(x) = 0
    [/tex]
     
  5. Jun 19, 2009 #4

    Mentallic

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    Aha

    [tex]tan(x+y)=\frac{sin(x+y)}{cos(x+y)}[/tex]

    But all I get using this result is

    [tex]tan(x+\frac{\pi}{2})=-cot(x)[/tex]

    It isn't helping just yet.
     
  6. Jun 19, 2009 #5

    Mentallic

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    Hang on...

    So the function now is [tex]f(x)=-\frac{x}{tan(x)}[/tex]

    and since the gradients of x and tanx at x=0 are equal, this gives it the value 1?
     
  7. Jun 19, 2009 #6

    arildno

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    Indeed.

    Or, as you can verify:
    [tex]x\tan(x+\frac{\pi}{2})=-\frac{x}{\sin(x)}\cos(x)[/tex]
     
  8. Jun 19, 2009 #7

    Mentallic

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    Well, I remember the result

    [tex]\lim_{x \to 0}\frac{x}{sin(x)}=1[/tex]

    and [itex]cos(0)=1[/itex] so I guess we can deduce that:

    [tex]\lim_{x \to 0}-\frac{x}{sin(x)}cos(x)=-1[/tex]

    However, I'm sure that the function doesn't exist at the point x=0, so if I were to draw the function, I would leave an empty circle at the point (0,-1)?

    Just like my previous mentioned function: [tex]f(x)=\frac{x+1}{x^2-1}[/tex]
    if I were to draw this function, I would quickly notice it is the same as [tex]f(x)=\frac{1}{x-1}[/tex] except [tex]x\neq -1[/tex]

    Can I do the same for [tex]f(x)=-x cot(x)[/tex] ? That is to say, can I find this equal to a simpler form (or more complicated if need be) of the same function, that instead is defined at x=0?
     
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