# Homework Help: Limits of modulus

1. Jun 1, 2014

### Portuga

Proof that $$\lim_{x→p} \frac {f(x)}{x - p} = 0 \Leftrightarrow \lim_{x \rightarrow p} \frac{f(x)}{\left | x - p\right |} = 0$$

Relevant equations
I think there is a useful theorem here, the conservation of signal:
Suppose that $\lim_{x\rightarrow p} f(x) = L$; $L > 0$; $\forall \varepsilon >0$, there is $\delta > 0$ so that $\forall x \in D_f$, $$0 < \left | x - p \right | < \delta \Rightarrow f(x) > 0$$.

The attempt at a solution
I've figured out a solution for the direct implication, and it's very easy:
$$\lim_{x\rightarrow p} \frac{f(x)}{x - p} = 0 \Rightarrow \forall \varepsilon >0 \exists \delta > 0 : 0 < \left | x - p \right | < \delta \Rightarrow \left | \frac{f(x)}{x - p} \right | < \varepsilon.$$
As $\left | x - p \right | = \left | \left | x - p \right | \right |$, the proof is over (at least, this is my opinion). The problem arise when I try to proof the reverse implication:
$$\lim_{x \rightarrow p} \frac{f(x)}{\left | x - p\right |} = 0 \Rightarrow \lim_{x→p} \frac {f(x)}{x - p} = 0 .$$
That's because I never considered elegant to use "reverse engeneering" to prove reverse implications (I mean, use the same strategy for both). For me, this is not elegant, and we must use another strategy to prove the reverse of the theorem, but I am with no ideas for this. Anyone has a clue of an independent method to prove it?

Last edited: Jun 1, 2014
2. Jun 1, 2014

### Portuga

Well, maybe a simple replacement of the symbol $\Rightarrow$ for $\Leftrightarrow$ in the proof above should solve the problem. I've seen it in some caculus textbooks. What do you think, guys?

Last edited: Jun 1, 2014
3. Jun 1, 2014

### pasmith

Hint: a two-sided limit exists if and only if both one-sided limits exist and are equal.