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$\text{Let}$$$\lim_{x\to d}h(x)=P,$$$$\lim_{x\to d}k(x)=Q\text{, and}$$$$h(x)\geq k(x)\text{ for all }x\text{ in an open interval containing }d.$$ $\text{Show}$$$P\geq Q$$

$\text{This is what I tried to do:}$$$|h(x)-P|\leq\epsilon,$$$$|k(x)-Q|\leq\epsilon,$$ $$-\epsilon\leq h(x)-P\leq\epsilon$$ $$-\epsilon\leq k(x)-Q\leq\epsilon$$ $$0\leq h(x)-k(x)-P+Q\leq0$$ $$|h(x)-k(x)-P+Q|=0\leq|h(x)-k(x)|+|P-Q|$$ $\text{But I get stuck. What am I doing wrong?}$

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The fifth line is wrong ... you multiplied the forth line with $-1$ and so have $\epsilon \ge Q - k(x) \ge -\epsilon$. Now everything works. AB, –  martini Mar 22 '12 at 17:46
    
You can also use the plain text instead of \text - that will save some ink –  Ilya Mar 22 '12 at 17:49

1 Answer 1

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You surely know that: $$\tag{1} f(x)\geq 0 \text{ in a neighbourhood of } d \text{ and } \lim_{x\to d}f(x)=L \quad \Rightarrow \quad L\geq 0\; ,$$ hence in order to solve your problem you can use (1) with $f(x):=h(x)-k(x)$ and $L:=P-Q$.

FWIW, the proof of (1) can be worked out by contradiction. In fact, assume that $L<0$ and take $\varepsilon = -\frac{L}{3}$ in the definition of limit: then you find that $\frac{4}{3}L< f(x)<\frac{2}{3}L<0$ in a suitable open neighbourhood $U$ of $d$; on the other hand, there exists an open neighbourhood $V$ of $d$ s.t. $f(x)\geq 0$ for $x\in V$; now in the open neighbourhood $W:=U\cap V$ your $f(x)$ has to be both $<0$ and $\geq 0$, which is a contradiction. Therefore $L\geq 0$.

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