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Look for a value of delta limit

  1. Mar 3, 2005 #1
    Hello, could someone please help me out with the following question?

    Q. Find the limit [tex]\mathop {\lim }\limits_{x \to 2} \left( {\frac{{2x - 1}}{{x + 1}}} \right)[/tex] and verify the result by [tex]\varepsilon - \delta [/tex] argument.

    I got the limit as 1 so I began to look for a value of delta.

    Required: [tex]\left| {\frac{{2x - 1}}{{x + 1}} - 1} \right| < \varepsilon [/tex] whenever [tex]0 < \left| {x - 2} \right| < \delta [/tex].

    [tex]\left| {\frac{{x - 2}}{{x + 1}}} \right| < \varepsilon[/tex] whenever [tex]0 < \left| {x - 2} \right| < \delta [/tex].

    I get stuck at this point. Even after I use division on the LHS I still don't get anywhere. Perhaps I need to consider values of x near 2 and deduce something from that? I dunno. Could someone please help me out? Any help would be great thanks.
     
  2. jcsd
  3. Mar 3, 2005 #2
    Note that |x - 2| = |x + 1 - 3| <= |x + 1| + |-3| <=> |x - 2| - 3 <= |x + 1| by the Triangle Inequality.

    --J
     
  4. Mar 3, 2005 #3
    Thanks for your help so far. Using the inequality I get:

    [tex]
    \left| {\frac{{x + 2}}{{x + 1}}} \right| \le \frac{{\left| {x + 2} \right|}}{{\left| {x + 2} \right| - 3}} \le \left| {x + 2} \right|
    [/tex]

    I then choose delta = epsilon. Is that step valid?
     
  5. Mar 3, 2005 #4
    No. I lead you a bit astray, actually. You'll note that your step is not valid, at, for instance, x = 1.1.

    Also, if you're making your denominator larger, it's actually making the total smaller, so the <= sign is not preserved by the first substitution.

    You'll find the other end of the triangle inequality much more useful.

    |x - 2| = |x + 1 - 3| >= |x + 1| - |-3| <=> |x + 1| <= |x - 2| + 3.

    Sorry about that.

    --J
     
    Last edited: Mar 3, 2005
  6. Mar 3, 2005 #5
    Ok thanks for the help. I'll give it another try.
     
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