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Practising limits for test

  1. Dec 18, 2004 #1
    Hi all, I've been practising limits for test, which is approaching very fast :) Let's have this limit:

    \lim_{x \rightarrow \frac{\pi}{2}_{+}} \frac{ \sqrt{1 - \sin x}}{\log \frac{2x}{ \pi }}

    I played with that so I got this form:

    \lim \frac{1}{\sqrt{1+\sin x}} \frac{1}{\frac{\log \frac{2x}{\pi}}{\cos x}}

    It can be seen that the second fraction is inversed well known limit

    \lim_{x \rightarrow 1} \frac{ \log{x} } { x - 1 } = 1

    Then, I can write

    = \lim \frac{1}{ \sqrt{1 + \sin x }} = \frac{ \sqrt{2} } { 2 }

    The problem is, I don't know how to justify my method exactly. You see, we have to base all your steps on the theorem we use for it, we must prove the conditions and so on...And in this limit I did it rather ty intuition...

    So how would the precise solution look like?

    Thank you.
    Last edited: Dec 18, 2004
  2. jcsd
  3. Dec 18, 2004 #2


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    Science Advisor

    I don't see any reason to make it so complex. AT x= π sin x= 0 so the numerator is 1. Assuming that that "3.14" in the denominator is actually π, then the denominator is log 2 which is not 0. Since all functions are continuous in the appropriate neighbohoods, the limit is 1/log(2).
  4. Dec 18, 2004 #3
    I'm sorry, I didn't write the LaTeX code properly, so I had to edit it. Now it's how it should be.
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