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Thread: Using Composition With Continuity

  1. #1
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    Using Composition With Continuity

    Let $\displaystyle g$ be defined on $\displaystyle \mathbb{R}$ by $\displaystyle g(1) := 0$, and $\displaystyle g(x) := 2$ if $\displaystyle x \neq 1$, and let $\displaystyle f(x) := x +1$ for all $\displaystyle x \in \mathbb{R}$. How would you show that $\displaystyle \lim_{x\to0}g \circ f \neq (g \circ f)(0)$ ?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    Let $\displaystyle g$ be defined on $\displaystyle \mathbb{R}$ by $\displaystyle g(1) := 0$, and $\displaystyle g(x) := 2$ if $\displaystyle x \neq 1$, and let $\displaystyle f(x) := x +1$ for all $\displaystyle x \in \mathbb{R}$. How would you show that $\displaystyle \lim_{x\to0}g \circ f \neq (g \circ f)(0)$ ?
    $\displaystyle g(f(0))=g(0+1)=g(1)=0$ but $\displaystyle g(f(x))=g(x+1)=2,\text{ }x\ne 0$...
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    Quote Originally Posted by Drexel28 View Post
    $\displaystyle g(f(0))=g(0+1)=g(1)=0$ but $\displaystyle g(f(x))=g(x+1)=2,\text{ }x\ne 0$...
    So to show they're not equal, should I prove that the limit of $\displaystyle \lim_{x\to0}g \circ f =2$ ?
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  4. #4
    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 $\displaystyle \lim_{x\to0}g \circ f =2$ ?
    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 $\displaystyle \lim_{x \to 0}g \circ f = 2$ so that I can show that the two are different, and I'm a bit stuck...

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

    I wanna show $\displaystyle |g(x+1)-0|=|g(x+1)|<\epsilon$ ...
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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 $\displaystyle \lim_{x \to 0}g \circ f = 2$ so that I can show that the two are different, and I'm a bit stuck...

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

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