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How to determine the x values where a function is continuous

  1. Mar 15, 2014 #1
    How would I find the x values for which a function is continuous ?, and how to tell whether it is a removable discontinuity, a jump discontinuity, or an infinite discontinuity ?

    Suppose the function is sqrt(9-x^2)/x^2-1
     
  2. jcsd
  3. Mar 15, 2014 #2
    [itex]\frac{\sqrt{9-x^2}}{x^2-1}=\frac{\sqrt{9-x^2}}{(x-1)(x+1)}[/itex]. So your function is not defined on the points 1 and -1, but it is continuous on the other points as a composition of basic continuous functions. To study the behavior of the function near the singularities, you calculate the limits, you will find [tex]\lim_{x\rightarrow 1} \frac{\sqrt{9-x^2}}{x^2-1}=+ \infty[/tex] from the right and [itex]-\infty[/itex] from the left. The opposite thing happens for -1.
     
  4. Mar 15, 2014 #3
    Wouldn't the numerator also have a restriction ?, the square root cant be negative. So would it be correct in saying that the graph is continuous on the intervals [-3,-1) U [3,1) ?
     
  5. Mar 15, 2014 #4

    arildno

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    f is continuous WHEREVER it is DEFINED (i.e, has a function value).
     
  6. Mar 15, 2014 #5

    Mark44

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    Interval notation goes left to right, so this would be written as [-3, -1) U (1, 3].
     
  7. Mar 15, 2014 #6

    Mark44

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    Technically, ##\lim_{x\to 1} \frac{\sqrt{9-x^2}}{x^2-1} ## does not exist, since the left and right limits are different (which is what you said). The notation that is often used is
    ##\lim_{x \to 1^+} \frac{\sqrt{9-x^2}}{x^2-1} = \infty## and ##\lim_{x \to 1^-} \frac{\sqrt{9-x^2}}{x^2-1} = -\infty##.
     
  8. Mar 15, 2014 #7

    arildno

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    The function is perfectly continuous at (-1,1) as well.
     
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