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Math Help - Functions question

  1. #1
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    Functions question

    The function f is defined by f(x)=sinx\ for -\frac{\pi}{4}<x\leq\frac{\pi}{4}, and\ f(x+\frac{\pi}{2})=f(x)+\sqrt{2} for\ all\ x \in \mathbb{R}

    1)\ Find\ an\ expression\ for\ f(x)\ for\ \frac{\pi}{4}<x\leq\frac{3\pi}{4}

    2)\ Prove\ that\ f\ is\ continuous\ at\ \frac{\pi}{4}. That\ is,\ prove\ \lim_{x\rightarrow\frac{\pi}{4}}f(x)=f(\frac{\pi}{  4})
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  2. #2
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    A possible solution is  f(x) = \frac{2\sqrt{2}x}{\pi}

    Since  \lim_{x \to \frac{\pi}{4}} \sin{x} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}},  \lim_{x \to \frac{\pi}{4}} \frac{2\sqrt{2}x}{\pi} = \frac{1}{\sqrt{2}} , f(x) is continuous at  \frac{\pi}{4}
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  3. #3
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    Quote Originally Posted by nahduma View Post
    A possible solution is  f(x) = \frac{2\sqrt{2}x}{\pi}

    Since  \lim_{x \to \frac{\pi}{4}} \sin{x} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}, \lim_{x \to \frac{\pi}{4}} \frac{2\sqrt{2}x}{\pi} = \frac{1}{\sqrt{2}} , f(x) is continuous at  \frac{\pi}{4}
    Shouldn't you consider the left and right hand limit?
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  4. #4
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    Quote Originally Posted by acevipa View Post
    Shouldn't you consider the left and right hand limit?
    Exactly.
    the right hand limit is the limit when  x>\frac{\pi}{4} ie. \lim_{x\to \frac{\pi}{4}} \frac{2\sqrt{2}x}{\pi}.
    the left hand limit is the limit when  x<\frac{\pi}{4} ie. \lim_{x\to\frac{\pi}{4}} \sin{x}.

    Hope this clears it up.
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