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Math Help - Using Composition With Continuity

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    Using Composition With Continuity

    Let g be defined on \mathbb{R} by g(1) := 0, and g(x) := 2 if x \neq 1, and let f(x) := x +1 for all x \in \mathbb{R}. How would you show that \lim_{x\to0}g \circ f \neq (g \circ f)(0) ?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    Let g be defined on \mathbb{R} by g(1) := 0, and g(x) := 2 if x \neq 1, and let f(x) := x +1 for all x \in \mathbb{R}. How would you show that \lim_{x\to0}g \circ f \neq (g \circ f)(0) ?
    g(f(0))=g(0+1)=g(1)=0 but g(f(x))=g(x+1)=2,\text{ }x\ne 0...
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    Quote Originally Posted by Drexel28 View Post
    g(f(0))=g(0+1)=g(1)=0 but g(f(x))=g(x+1)=2,\text{ }x\ne 0...
    So to show they're not equal, should I prove that the limit of \lim_{x\to0}g \circ f =2 ?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    So to show they're not equal, should I prove that the limit of \lim_{x\to0}g \circ f =2 ?
    Is there a need to prove it?
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    Quote Originally Posted by Drexel28 View Post
    Is there a need to prove it?
    Yea, I'm trying to prove that \lim_{x \to 0}g \circ f = 2 so that I can show that the two are different, and I'm a bit stuck...

    so \lim_{x \to 0}g \circ f = \lim_{x \to 0}g(x+1)

    I wanna show |g(x+1)-0|=|g(x+1)|<\epsilon ...
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  6. #6
    Member mabruka's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    Yea, I'm trying to prove that \lim_{x \to 0}g \circ f = 2 so that I can show that the two are different, and I'm a bit stuck...

    so \lim_{x \to 0}g \circ f = \lim_{x \to 0}g(x+1)

    I wanna show |g(x+1)-0|=|g(x+1)|<\epsilon ...
    i dont follow :S
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