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Math Help - Left hand limit

  1. #1
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    Left hand limit

    Since I haven't learned to use LaTex yet, for convenience, let L.H.f(x) be the limit of f as x approaches c from the left.

    1.) If f(x)>0 and L.H.f(x) = 0, show that L.H.(1/f(x)) = infinity.
    2.) Give an example of a function f for which L.H.f(x) = 0, but L.H.(1/f(x)) is neither infinity nor negative infinity.
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  2. #2
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    Quote Originally Posted by spoon737 View Post
    Since I haven't learned to use LaTex yet, for convenience, let L.H.f(x) be the limit of f as x approaches c from the left.

    1.) If f(x)>0 and L.H.f(x) = 0, show that L.H.(1/f(x)) = infinity.
    2.) Give an example of a function f for which L.H.f(x) = 0, but L.H.(1/f(x)) is neither infinity nor negative infinity.
    Hints: function is positive, but it goes to zero as x goes from negative infinity to c, so it must be a decreasing function in that domain. Therefore, 1/f(x) will go to infinity. To show this, all you need to do is use the basic properties of limits.

    lim_{ x\rightarrow a} \frac{f(x)}{g(x)} = \frac{lim_{ x\rightarrow a} f(x)}{lim_{x\rightarrow a} g(x)}

    In this case (#1):

    lim_{ x\rightarrow a} \frac{1}{f(x)} = \frac{lim_{ x\rightarrow a} 1}{lim _{x\rightarrow a} f(x)} \Rightarrow \frac{1}{0} = \infty
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  3. #3
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    Thanks for your quick response. I was thinking the proof would be similar to yours, but I wasn't sure if that reasoning alone would be rigorous enough (this is for an analysis class). I was thinking I needed to use the epsilon-delta definition in my proof.
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    Quote Originally Posted by spoon737 View Post
    Since I haven't learned to use LaTex yet, for convenience, let L.H.f(x) be the limit of f as x approaches c from the left.

    1.) If f(x)>0 and L.H.f(x) = 0, show that L.H.(1/f(x)) = infinity.
    If \lim_{x\to c} f(x) = \infty it means for any N>0 there exists a X> 0 such that if x>X then f(x) > N. This means that 0< 1/f(x) < 1/N. Thus, |1/f(x)| < 1/N for x>X. Let \epsilon > 0 then we can choose N so large that 1/N < \epsilon and so |1/f(x)| < \epsilon. Thus, if \lim_{x\to c}f(x) = \infty for f(x) > 0 it means \lim_{x\to c}1/f(x) = 0. Can you imitate this proof to prove what you are trying to show?
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  5. #5
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If \lim_{x\to c} f(x) = \infty it means for any N>0 there exists a X> 0 such that if x>X then f(x) > N.
    That is for when f(x) tends to infinity as x itself tends to infinity. This question is slightly different: we say that f(x) tends to infinity as x tends to c from the left iff for any N > 0, there exists a δ > 0 such that if c−δ < x < c then f(x) > N.
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  6. #6
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    Given f(x) = \left\{ {\begin{array}{rl}<br />
   { - x} & {x \in Q^ -  }  \\<br />
   x & {x \in \Re ^ -  \backslash Q}  \\<br />
\end{array}} \right.
    Then what are these \lim _{x \to 0^ -  } f(x)\,\& \,\lim _{x \to 0^ -  } \frac{1}{{f(x)}} ?
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  7. #7
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    Quote Originally Posted by JaneBennet View Post
    That is for when f(x) tends to infinity as x itself tends to infinity. This question is slightly different: we say that f(x) tends to infinity as x tends to c from the left iff for any N > 0, there exists a δ > 0 such that if c−δ < x < c then f(x) > N.
    I know. I was doing a different problem because I did not realize he asked that.
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