# How to prove that $\lim_{x\rightarrow0}\frac{1}{2x+1}=1$ using $\delta-\epsilon$ definition?

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

• @obscurans Thanks for you effort to edit posts into better shape. I'll just mention that you can use \lim to get nicer rendering and also proper spacing for the limit symbol. For example $\lim_{n\to\infty} x_n$ $\lim_{n\to\infty} x_n$ or $\lim\limits_{n\to\infty} x_n$ $\lim\limits_{n\to\infty} x_n$. – Martin Sleziak Dec 19 '18 at 5:41

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.

$$\mid\frac{-2x}{2x+1}\mid=\mid\frac{-2}{2+\frac1x}\mid\lt\mid\frac2{\frac 1x}\mid\lt\epsilon$$. So make $$\mid x\mid\lt\frac{\epsilon}2$$. That is, $$\delta =\frac {\epsilon}2$$.

For $$\quad\delta<\dfrac{\varepsilon}{2(1-\varepsilon)}\implies\quad\displaystyle\left|\dfrac{-2x}{2x+1}\right|<\dfrac{2\delta}{2\delta+1}=\dfrac{1}{1+\frac{1}{2\delta}}<\varepsilon$$