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Thread: Limit of a function with trigonometric functions

  1. #1
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    Limit of a function with trigonometric functions

    Hi everyone,

    I am having trouble calculating the following limit

    \lim\limits_{x \rightarrow 0}{\frac{\sqrt{1+\sin^2{x}}-\sqrt{1-\sin^2{x}}}{1-\cos{x}}}

    I know that 1-\sin^2{x} is equal to \cos^2{x}, but what do I do with 1+\sin^2{x}?

    I would appreciate any help.
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  2. #2
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    Quote Originally Posted by thomasdotnet View Post
    Hi everyone,

    I am having trouble calculating the following limit

    \lim\limits_{x \rightarrow 0}{\frac{\sqrt{1+\sin^2{x}}-\sqrt{1-\sin^2{x}}}{1-\cos{x}}}

    I know that 1-\sin^2{x} is equal to \cos^2{x}, but what do I do with 1+\sin^2{x}?
    Multiply by
    {\frac{\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}}}{\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}}}}
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  3. #3
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    do you know what l'hospital's rule for computing limits is?
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  4. #4
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    Thank you very much guys for your responses!

    Using l'Hospital's rule would of course be possible, but it does render the numerator of the fraction a bit ugly, thus I don't think that using l'Hospital's rule was the original intention.

    I did the expansion by {\frac{\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}}}{\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}}}} and simplified the numerator, but I have to admit that I got stuck again at this point:
    {\frac{2\sin^2{x}}{(1-\cos{x})(\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}})}}
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  5. #5
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    Quote Originally Posted by thomasdotnet View Post
    Thank you very much guys for your responses!

    Using l'Hospital's rule would of course be possible, but it does render the numerator of the fraction a bit ugly, thus I don't think that using l'Hospital's rule was the original intention.

    I did the expansion by {\frac{\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}}}{\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}}}} and simplified the numerator, but I have to admit that I got stuck again at this point:
    {\frac{2\sin^2{x}}{(1-\cos{x})(\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}})}}
    {\frac{2\sin^2{x}}{(1-\cos{x})(\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}})}}= {\frac{2(1-\cos^2{x})}{(1-\cos{x})(\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}})}} ={\frac{2[(1-\cos{x})(1+\cos{x})]}{(1-\cos{x})(\sqrt{1+\sin^2{x}}+\sqrt{1-\sin^2{x}})}}
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  6. #6
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    That solves the problem, thanks!
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