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Math Help - More Functional Limits

  1. #1
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    More Functional Limits

    (1) Let g: A \rightarrow \mathbb{R} and assume that f is a bounded function on A \subseteq \mathbb{R} (i.e., there exists M > 0 satisfying |f(x)| \leq M for all x \in A). Show that if lim_{x \rightarrow c} g(x)=0, then lim_{x \rightarrow c} g(x)f(x)=0 as well.


    (2) Assume f(x) \geq g(x) for all x in some set A on which f and g are defined. Show that for any limit point c of A we must have lim_{x \rightarrow c} f(x) \geq lim_{x \rightarrow c} g(x).


    Thanks for looking and any help I can be provided! I am really unsure of how to approach these at all.
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  2. #2
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    Quote Originally Posted by shadow_2145 View Post
    (1) Let g: A \rightarrow \mathbb{R} and assume that f is a bounded function on A \subseteq \mathbb{R} (i.e., there exists M > 0 satisfying |f(x)| \leq M for all x \in A). Show that if lim_{x \rightarrow c} g(x)=0, then lim_{x \rightarrow c} g(x)f(x)=0 as well.
    Give condition implies:

    \forall \epsilon > 0 ,\exists \delta > 0: |x - c| < \delta \Rightarrow |g(x) - 0| < \epsilon

    That means if we choose \epsilon as \frac{\epsilon}{M}, there exists a delta such that |x - c| < \delta \Rightarrow |g(x) - 0| < \frac{\epsilon}{M}.

    Choose this delta to prove continuity of f(x)g(x)...

    More precisely, we have to prove that:

    \forall \epsilon > 0 ,\exists \delta > 0: |x - c| < \delta \Rightarrow |f(x)g(x) - 0| < \epsilon

    For any \epsilon choose the \delta said previously,

    Now |f(x)g(x) - 0| = |f(x)g(x)| <M\frac{\epsilon}{M}= \epsilon
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  3. #3
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    Perfect! Thanks!
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  4. #4
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    Quote Originally Posted by shadow_2145 View Post
    (2) Assume f(x) \geq g(x) for all x in some set A on which f and g are defined. Show that for any limit point c of A we must have lim_{x \rightarrow c} f(x) \geq lim_{x \rightarrow c} g(x).
    Assume the contrary, that's \lim_{x\rightarrow{c}}f(x)<\lim_{x\rightarrow{c}}g  (x)

    Or equivalently: 0<\lim_{x\rightarrow{c}}[g(x)-f(x)]=d

    By definition: <br />
\forall \varepsilon  > 0,\exists \delta _\varepsilon   > 0/{\text{ if }}0 < \left| {x - c} \right| < \delta _\varepsilon  {\text{ we have:}}{\text{ }}\left| {g\left( x \right) - f\left( x \right) - d} \right| < \varepsilon <br />

    Choose: <br />
\varepsilon  = \tfrac{d}<br />
{2}<br />
thus we have: <br />
{\text{if }}0 < \left| {x - c} \right| < \delta _{\tfrac{d}<br />
{2}}  \Rightarrow 0 < \tfrac{1}<br />
{2} \cdot d < g\left( x \right) - f\left( x \right) < \tfrac{3}<br />
{2} \cdot d<br />

    And this is absurd since it goes against the hypothesis <br />
f\left( x \right) > g\left( x \right)<br />
for all <br />
x \in A<br />
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