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Limits and Continuous Functions problem

  1. Jun 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Define the function at a so as to make it continuous at a.
    [tex]f(x)=\frac{4-x}{2-\sqrt{x}}; a = 4[/tex]


    2. Relevant equations
    [tex]\lim_{x \rightarrow 4} \frac{4-x}{2-\sqrt{x}}[/tex]


    3. The attempt at a solution
    I cannot think of how to manipulate the denominator to achieve f(4), so I start by finding the left and right hand limits then applying the informal definition of a limit.

    [tex]\lim_{x \rightarrow +4}\frac{4-x}{2-\sqrt{x}} = 4[/tex]
    [tex]\lim_{x \rightarrow -4}\frac{4-x}{2-\sqrt{x}} = 4[/tex]

    I do not know if this is the right way to solve this problem though. If I use this method, I am still left with a function where f(4) is discontinuous. Any hints would be greatly appreciated.
     
  2. jcsd
  3. Jun 18, 2011 #2

    eumyang

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    Homework Helper

    Use the difference of two squares pattern
    [tex]a^2 - b^2 = (a - b)(a + b)[/tex]
    where a2 - b2 corresponds to 4 - x. See what happens.
     
  4. Jun 18, 2011 #3
    Wow that seems so obvious now. I completely overlooked it. Thank you so much for the help.
     
  5. Jun 18, 2011 #4

    Mark44

    Staff: Mentor

    Your notation needs a bit of tweaking. What you think you wrote, and what you actually wrote are two different things. Your limits were supposed to be one-sided limits around 4. The second limit, above, was as x approaches -4, which is not in the domain of the square root function.

    This is what you should have written:
    [tex]\lim_{x \rightarrow 4^+}\frac{4-x}{2-\sqrt{x}} = 4[/tex]
    [tex]\lim_{x \rightarrow 4^-}\frac{4-x}{2-\sqrt{x}} = 4[/tex]

    Notice that the plus and minus signs appear after 4.
     
  6. Jun 19, 2011 #5
    Can you use L'Hospital's rule for something like this? The limit is in the indeterminate form of 0/0 and the result of one iteration of L'Hospital is 2√x which is continuous at x=4... Am I cheating here?
     
  7. Jun 19, 2011 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, you could use L'Hopital's rule but that would be "over-kill". eumyang's suggestion is much simpler.
     
  8. Jun 19, 2011 #7
    I guess it doesn't seem obvious to me using the difference of squares then... Would you mind writing it out?

    Ken
     
  9. Jun 19, 2011 #8

    gb7nash

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    Homework Helper

    [tex]4 - x = (2 - \sqrt{x})(2 + \sqrt{x})[/tex]

    After that, you get cancellation.
     
  10. Jun 19, 2011 #9
    Omg! I was working the pants off the denominator and totally missed that the numerator was the key! Thanks!
    Ken
     
  11. Jun 20, 2011 #10

    HallsofIvy

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    Another, similar, way to do this is to multiply both numerator and denominator by [itex]2+ \sqrt{x}[/itex].
     
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