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LIMITS, complicated square roots and factoring

  1. Sep 16, 2008 #1
    The question is as follows:
    [tex]\frac{lim}{h\rightarrow0}[/tex] [tex]\frac{\sqrt{1+h}-1}{h}[/tex]

    I don't know if the way I approached the question is right, I'll give you a step by step of what I attempted:
    First I converted the square root into 11/2 and h1/2 (can I do that? Is that correct?)
    Then I continued to evaluate, getting h1/2 over h
    Finally I determined the value to be 0 because 0-1/2 is 0.

    I feel as though my first step is where I might have errored but I'm not sure how else to approach it, perhaps by converting the h to a h-1
  2. jcsd
  3. Sep 16, 2008 #2


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    Are you saying that [itex]\sqrt{1 + h} = \sqrt{1} + \sqrt{h}[/itex] (for all h)? If so, then no, this is not correct.

    Hint: Multiply top and bottom by [itex]\sqrt{1+h} + 1[/itex], and see what happens.
  4. Sep 17, 2008 #3


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    Hi again Susan_t;

    When you do the multiplication suggested above, think about how the product [tex] (a-b)(a+b) [/tex] is expanded.

    I don't like the word "trick" in relation to mathematics, but the suggested step given by morphism will be a handy one to master in your studies.
  5. Sep 17, 2008 #4
    Yes! that definitely got my wheels turning, I ended up rationalizing it by the multiplying of both the top and the bottom of the equation with the opposite of the numerator and an answer of one half. Thanks for the help again
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