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Math Help - Find the limit of a function

  1. #1
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    Find the limit of a function

    can you help me solving this question
    without using L'HOPITAL'S RULE

    Find the limit of a function-lin1.jpg
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  2. #2
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    just put t=x-\frac\pi4.
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  3. #3
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    If you know anything about derivatives ...

    In general: f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}

    Imagine: a = \tfrac{\pi}{4} and f(x) = \cos^2 (x)
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    just put t=x-\frac\pi4.
    OK , I put t=x-\frac\pi4
    I got the following restul !!


    Attached Thumbnails Attached Thumbnails Find the limit of a function-lim8.jpg  
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by razemsoft21 View Post
    OK , I put t=x-\frac\pi4
    I got the following restul !!


    use the identity sin2x=2cos^2x-1

    and then the limit \frac{sinx}{x}
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  6. #6
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    Quote Originally Posted by o_O View Post
    If you know anything about derivatives ...

    In general: f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}

    Imagine: a = \tfrac{\pi}{4} and f(x) = \cos^2 (x)
    Since the question is posted in PRE-Calculus it might be IF rather than if ....
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  7. #7
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    Quote Originally Posted by razemsoft21 View Post
    OK , I put t=x-\frac\pi4
    I got the following restul !!


    Use cos(t+ \pi/4)= cos(t)cos(\pi/4)- sin(t)sin(\pi/4) = (\sqrt{2}/2)cos(t)- (\sqrt{2}/2)sin(t) so cos^2(t+ \pi/4)= \frac{1}{2}cos^2(t)- sin(t)cos(t)+ \frac{1}{2}sin^2(t)

    \frac{cos^2(t+ \pi/4)- \frac{1}{2}}{t}= \frac{1}{2}\frac{cos^2(t)-1}{t}- \frac{sin(t)}{t}cos(t)+ \frac{1}{2}\frac{sin(t)}{t}sin(t) = -\frac{1}{2}\frac{sin(t)}{t}sin(t)- \frac{sin(t)}{t}cos(t)+\frac{1}{2}\frac{sin(t)}{t}  sin(t) = -\frac{sin(t)}{t}cos(t)
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  8. #8
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    Thumbs up Thanks

    Quote Originally Posted by o_O View Post
    If you know anything about derivatives ...

    In general: f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}

    Imagine: a = \tfrac{\pi}{4} and f(x) = \cos^2 (x)
    Thank you for help, it works: razemsoft21 o_O

    definition of f'(pi/4) = -2cos(pi/4).sin(pi/4) = -1
    the same result when I use L'HOPITAL'S RULE
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