Proving limit using epislon definition

Question

How can I use $\epsilon-\delta $ to prove the following?

If $mh \le k \le Mh$ $ (m,M > 0)$ ,

$\lim_{h\to 0}\frac{G(h)}{h}=0\iff lim_{h\to 0}\frac{G(h)}{k}=0$.

Answer 1

The "If" Part:

If $\lim_{h\to 0}\frac{G(h)}{h}=0$, the for all $\epsilon>0$, there exists a $\delta>0$, such that

$$\left|\frac{G(h)}{h}\right|<\epsilon$$

whenever $0<|h|<\delta$.

But then we have

$$\left|\frac{G(h)}{k}\right|\le \left|\frac{G(h)}{mh}\right|=\frac1m\left|\frac{G(h)}{h}\right|<\epsilon/m$$

whenever $0<|h|<\delta$, which implies $\lim_{h\to 0}\frac{G(h)}{k}=0$.

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The "Only If" Part:

If $\lim_{k\to 0}\frac{G(h)}{k}=0$, the for all $\epsilon>0$, there exists a $\delta>0$, such that

$$\left|\frac{G(h)}{k}\right|<\epsilon$$

whenever $0<|h|<\delta$.

But then we have

$$\left|\frac{G(h)}{k}\right|\le \left|\frac{G(h)}{Mh}\right|=\frac1M\left|\frac{G(h)}{h}\right|<\epsilon/M$$

whenever $0<|h|<\delta$, which implies $\lim_{h\to 0}\frac{G(h)}{h}=0$.

And we are done!