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Use the Intermediate Value Theorem

  1. Sep 27, 2005 #1
    Hello everyone,

    Unfortunately, my calculus class has just gotten to my least favorite subject in the world.....limits. There are a few questions I am hopelessly stuck on. Any help would be appreciated.

    The limit as x approaches 1 of (square root x -x^2)/(1-square root x)

    I have no idea how to go about this one. I know that usually when you have a square root you rationalize the equation, but when you have one in the denominator and numerator what do you do?

    Use the Intermediate Value Theorem toshow that there is a root of the given equation in the specified interval. Tanx=2x, (0,1.4)

    For this one I changed it into the form f(x)=tanx-2x , then I found that f(0)=0 and f(1.4)=2.99 therefore f(0) < 0 < f(2). Is this all I have to do, Is this enough to satisfy the question asking if there is a root between (0,1.4)???

    My next questions are ones I was able to get answers for but am unsure of whether I did them correctly.

    Find the limit as x->2 of (2x^2+1)/(x^2+6x-4).

    For this one I believe you are able to simply substitute 2 in for x and that will give you an answer of 3/4

    Find the lim as x ->1 of (x^3-1)/(x^2-1)

    For this question I factored out (x-1) and from there I was able to get an answer of (3/2).

    My other questions are similar, and if I know I did these right I can be fairly sure the others are correct. Thank you so much for your time and help.
  2. jcsd
  3. Sep 27, 2005 #2
    For the first question have you tried L'Hopitals rule?
  4. Sep 28, 2005 #3
    I'm not aware of that rule, could you elaborate on it? Everytime I try the question I get that the limit is undefined, could it be possible that the limit just does not exist?
  5. Sep 28, 2005 #4


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    Homework Helper

    In #1,
    [tex]\lim_{x \rightarrow 1} \frac{\sqrt{x - x ^ 2}}{1 - \sqrt{x}}[/tex]
    You can get rid of the square root in the denominator by multiplying both numerator and denominator with [itex]1 + \sqrt{x}[/itex], so the denominator will become 1 - x. Just do it and see what you get.
    Or you can try:
    [tex]\lim_{x \rightarrow 1} \frac{\sqrt{x - x ^ 2}}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \frac{\sqrt{x}\sqrt{1 - x}}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \frac{\sqrt{x}\sqrt{(1 - \sqrt{x})(1 + \sqrt{x})}}{1 - \sqrt{x}}[/tex]
    Can you go from here?
    Viet Dao,
  6. Sep 28, 2005 #5


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    Homework Helper

    That's possible but not necessarily so.

    L'Hopital's rule says that when you encounter the undeterminate forms 0/0 or [itex]\infty /\infty[/itex], you're allowed to take (separately!) the derivative of nominator and denominator. So if the limit f(x)/g(x) yields such a form, you may replace it by f'(x)/g'(x).
  7. Sep 28, 2005 #6
    Thanks guys I think I got it from here!!! I really appreciate it!
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