Basic limit question to understand the methods

Question

I have a very basic question about proving limits with the epsilon-delta method.

So i want to prove $\lim _{x\to 0}\left(\frac{1}{1-2x}\right)\:=\:1$ . first, i write it like that:
$\left|\frac{1}{1-2x}\:-\:1\right|\:$ = $\left|\frac{2x}{1-2x}\right|\:$ = $\frac{2\left|x\right|}{\left|1-2x\right|}\:$ and from here i stuck. I know that i need to get |x|< some expression of epsilon and stuff that will be actually my delta. What i need to do from here?

Answer 1

We want to show that $\dfrac{2}{|1-2x|}|x|$ can be made small (less than a preassigned $\epsilon$) by taking $|x|$ suitably close to $0$.

We can control $|x|$. The issue is with the term $\frac{2}{|1-2x|}$, which could potentially spoil things: We must make sure that $1-2x$ stays well away from $0$.

If we first of all specify that $\delta\lt \frac{1}{4}$, that forces $\frac{1}{2}\lt 1-2x$, which makes $\frac{2}{|1-2x|}\lt 4$.

Now it is easy to specify a $\delta$ that will work. We make $\delta=\min(\frac{1}{4},\frac{\epsilon}{4})$.

Answer 2

Let $\varepsilon>0$, then when $|x|<\max{\frac{\varepsilon}{4},\frac{1}{4}}$ you have:

$$\frac{2|x|}{|1-2x|}<\frac{2|x|}{\frac{1}{2}}=4|x|$$

It's because $|x|<\frac{1}{4}$, so $|1-2x|>\frac{1}{2}$, and finally:

$$4|x|<\varepsilon$$

Because $|x|<\frac{\varepsilon}{4}$.