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I know

  1. If $f:D(\subset\mathbb R)\to\mathbb R,c$ is a limit point of $D,$ and $f(x)\ge(\text{resp.}\le)~a~\forall~x\in D-\{c\},$ then $\displaystyle\lim_{x\to c}f(x)\ge(\text{resp.}\le)~a.$ (Provided the limit exists)

  2. If $f,g,h:D(\subset\mathbb R)\to \mathbb R,c$ is a limit point of $D,$ and $f(x)\le g(x)\le h(x)~\forall~x\in D-\{c\},$ then $\displaystyle\lim_{x\to c}f(x)=\displaystyle\lim_{x\to c}h(x)\implies \displaystyle\lim_{x\to c}f(x)=\displaystyle\lim_{x\to c}g(x)=\displaystyle\lim_{x\to c}h(x).$

I wounder whether the following result holds:

$\diamond$ Let $f,g:D(\subset\mathbb R)\to\mathbb R,c$ is a limit point of $D,f(x)\ge g(x)~\forall~x\in D-\{c\}.$ Do all these imply $\displaystyle\lim_{x\to c}f(x)\ge\displaystyle\lim_{x\to c}g(x)?$ (Provided the limits exist)

For then it will be more powerful result from which both 1 and 2 follow.

My Attempt:

$F-\epsilon<f(x)<F+\epsilon~\forall~x\in D\cap N'(\delta_1)$

$G-\epsilon<g(x)<G+\epsilon~\forall~x\in D\cap N'(\delta_1)$

Set $\delta=\min\{\delta_1,\delta_2\}.$ Then $G-\epsilon<g(x)\le f(x)<F+\epsilon~\forall~x\in D\cap N'(\delta)$ i.e. $G<F+\epsilon.$

Since $\epsilon>0$ is arbitrary so $G\le\displaystyle\inf_{\epsilon>0}(F+\epsilon)=F.$

Is my attempt correct?

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1 Answer 1

up vote 1 down vote accepted

Basically, you can just use 1. for the function $h = f - g$ and $a=0$.

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ohhhhhhhh!!!! nooooooooooo........:( –  Sriti Mallick May 25 '13 at 12:43
    
So I simply need one result: $f:D\to\mathbb R$ and $c$ is a limit point od $D.$ Then $f\ge0$ on $D-\{c\}\implies\displaystyle\lim_{x\to c}{f(x)}\ge0.$ Right? –  Sriti Mallick May 25 '13 at 12:47
    
@SritiMallick, yes, exactly. –  xen May 25 '13 at 13:07
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