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Thread: epsilon-delta proof

  1. #1
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    epsilon-delta proof

    $\displaystyle \frac{x+1}{x-2}\rightarrow\frac{1}{4}$ as$\displaystyle x\rightarrow -2$

    I've got that $\displaystyle \frac{3}{4}\mid\frac{x+2}{x-2}\mid<\epsilon$

    and obviously $\displaystyle x\rightarrow-2$ implies $\displaystyle 0< x+2<\delta$
    (modulus signs around x+2 in the last bit)

    But the problem is getting the modulus of 1/(x-2) between something to get a value for delta in terms of epsilon.

    Any help greatly appreciated.
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  2. #2
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    Quote Originally Posted by featherbox View Post
    $\displaystyle \frac{x+1}{x-2}\rightarrow\frac{1}{4}$ as$\displaystyle x\rightarrow -2$

    I've got that $\displaystyle \frac{3}{4}\mid\frac{x+2}{x-2}\mid<\epsilon$

    and obviously $\displaystyle x\rightarrow-2$ implies $\displaystyle 0< x+2<\delta$
    (modulus signs around x+2 in the last bit)

    But the problem is getting the modulus of 1/(x-2) between something to get a value for delta in terms of epsilon.
    Start with $\displaystyle |x+2|<1$ then you can get $\displaystyle \left|\frac{1}{2-x}\right|<\frac{1}{3}$.

    Let $\displaystyle \delta =\min\left\{1,\epsilon\right\}$.
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