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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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    You can also use the plain text instead of `\text` - that will save some ink2012-03-22

1 Answers 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 $L\geq 0$.