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Continuous function.

  1. Apr 19, 2004 #1
    Im having a little trouble with this question.

    If f is continuous at [tex]c[/tex] and [tex]f(c) < 5[/tex], prove that there exists a [tex]\delta > 0[/tex] such that [tex]f(x) < 7[/tex] for all [tex]x \in (c - \delta , c + \delta)[/tex]

    So we are given that f is continuous at c.
    So [tex]\lim_{x \to c}f(x) = f(c) < 5[/tex]
    [tex]\forall \epsilon > 0 \exists \delta > 0[/tex] such that whenever [tex]|x - c| < \delta[/tex] then [tex]|f(x) - f(c)| < \epsilon[/tex]

    [tex]|x - c| < \delta[/tex]
    [tex]-\delta < x - c < \delta[/tex]
    [tex]c - \delta < x < c + \delta[/tex]

    Ok now im getting lost..
    I know i have to do something with [tex]|f(x) - f(c)| < \epsilon[/tex]
    [tex]|f(x) - 5| < \epsilon[/tex]
    [tex]-\epsilon < f(x) - 5 < \epsilon[/tex]
    [tex]5 - \epsilon < f(x) < 5 + \epsilon[/tex]
    we want [tex]f(x) < 7[/tex] so .... am i on the right track??? How can i find the [tex]\delta > 0[/tex] that satisfies this?
  2. jcsd
  3. Apr 19, 2004 #2
    Yes, you're on the right track. Take [itex]\epsilon=2[/itex]. From the limit you mentioned, we know that there exists a [itex]\delta>0[/itex] such that [itex]3<f(x)<7[/itex] and so you are done. The question was asking you to show that a delta exists, and the limit demonstrates that it clearly does, so you are done.
  4. Apr 19, 2004 #3
    I was actually thinking about making [tex]\epsilon = 2[/tex] but thought it was to easy. Ok so thanks for explaining the question a bit more to me, i understand it now. :smile:
  5. Apr 19, 2004 #4

    Isn't what you're asked to prove equivalent to,

    [tex]\lim_{x \to c}f(x) = f(c) < 7[/tex]

    which follows trivially from,

    [tex]\lim_{x \to c}f(x) = f(c) < 5[/tex]

    which was what you were given?
  6. Apr 20, 2004 #5


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    No, that's not what was given. Given that the limit is less than 5 it is true, but requires proof, that, for x close to c, f(x)< 5. gimpy was asked to prove the slightly simpler case: that, for x close to c, f(x)< 7.
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