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Thread: Prove Limits Help

  1. #1
    Junior Member
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    Prove Limits Help

    $\displaystyle \lim_{x\rightarrow 1} \frac{2+4x}{3} = 2 $

    $\displaystyle 0<\left | x-1 \right | < \delta $


    $\displaystyle \left | \frac{2+4x}{3}-2 \right | <\epsilon $


    simplified i get $\displaystyle \left | x-4 \right |< \frac{\epsilon }{4} $

    i set $\displaystyle \frac{\epsilon }{4} = \delta $
    then i get $\displaystyle 0< \left | x-1 \right | < \frac{\epsilon }{4} $

    I dont have an answer key for that problem in my calculus book so i was wondering if i did it right?
    Thankss
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  2. #2
    MHF Contributor
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    Re: Prove Limits Help

    You want to prove $\displaystyle \left | \frac{2+4x}{3}-2 \right | <\epsilon $.

    $\displaystyle \left | \frac{4x-4}{3} \right | <\epsilon $

    $\displaystyle \left | \frac{4}{3} (x-1)\right | <\epsilon $

    $\displaystyle | x-1 | <\frac{3}{4}\,\epsilon $. So it looks like it would work to set $\displaystyle \delta=\frac{3}{4}\,\epsilon $.

    Now you need to write out the proof....

    There's a really good paper describing how to construct an epsilon-delta limit proof - it's in one of the "sticky" threads in this forum.

    - Hollywood
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