How to prove that $\lim\limits_{x\rightarrow 0}\frac{1}{2x+1}=1$ using $\delta-\epsilon$ definition?
I have the following so far-
$|x-0|<\delta$ and $\left|\frac{1}{2x+1}-1\right|<\epsilon$
which boils to $\left|\frac{-2x}{2x+1}\right|<\epsilon$
I'm trying to figure out a way to factor out an $x$ but don't know how
On
$\mid\frac{-2x}{2x+1}\mid=\mid\frac{-2}{2+\frac1x}\mid\lt\mid\frac2{\frac 1x}\mid\lt\epsilon$. So make $\mid x\mid\lt\frac{\epsilon}2$. That is, $\delta =\frac {\epsilon}2$.
On
So you want $|\dfrac{2x}{2x+1}|<\epsilon $.
There are two ways go here.
For the first, note that you don't need the best bounds, just one that will work.
Soppose $|x| < \dfrac14$. Then $|2x| < \dfrac12$ so $|1+2x| > 1-2\cdot \dfrac14 = \dfrac12$. Therefore $|\dfrac{2x}{2x+1}| \lt |\dfrac{2x}{\frac12}| =|4x| $. To make $|4x| < \epsilon$, you just need $|x| < \dfrac{\epsilon}{4}$.
Therefore, if $|x| < \min(\dfrac14, \dfrac{\epsilon}{4})$, we have $|\dfrac{2x}{2x+1}|<\epsilon $.
If you want the best bounds, $|\dfrac{2x}{2x+1}|<\epsilon $ is the same as $\dfrac1{\epsilon} \lt |\dfrac{2x+1}{2x}| = |1+\dfrac{1}{2x}| $.
If $x > 0$, $|1+\dfrac{1}{2x}| =1+\dfrac{1}{2x} $ so we want $\dfrac1{\epsilon} \lt 1+\dfrac{1}{2x} $ or $\dfrac1{\epsilon}-1 \lt \dfrac{1}{2x} $ or, assuming $\epsilon < 1$, $2x \lt \dfrac1{\dfrac1{\epsilon}-1} = \dfrac{\epsilon}{1-\epsilon} $ or $x \lt\dfrac{\epsilon}{2(1-\epsilon)} $.
If $x < 0$, to make $\dfrac1{\epsilon} \lt |1+\dfrac{1}{2x}| $ be easy to work with, we need $1+\dfrac{1}{2x} < 0$ or $x > -\dfrac12$.
If this holds, then $|1+\dfrac{1}{2x}| =-\dfrac{1}{2x}-1 $, so we want $-\dfrac{1}{2x}-1 \gt \dfrac1{\epsilon} $ or $-\dfrac{1}{2x} \gt \dfrac1{\epsilon}+1 $ or $-2x \lt \dfrac1{\dfrac1{\epsilon}+1} $ or $-x \lt \dfrac{\epsilon}{2(1+\epsilon)} $.
I always prefer to get a simple bound which is not the best.
For $\quad\delta<\dfrac{\varepsilon}{2(1-\varepsilon)}\implies\quad\displaystyle\left|\dfrac{-2x}{2x+1}\right|<\dfrac{2\delta}{2\delta+1}=\dfrac{1}{1+\frac{1}{2\delta}}<\varepsilon$