Use the $\varepsilon$ - $\delta$ definition to prove $\lim_{x\to\,-1}\frac{x}{2x+1}=1$ Use the $\varepsilon$ - $\delta$ definition of limit to prove that $\displaystyle\lim_{x\to\,-1}\frac{x}{2x+1}=1$.
My working:
$\left|\frac{x}{2x+1}-1\right|=\left|\frac{-x-1}{2x+1}\right|=\frac{1}{\left|2x+1\right|}\cdot \left|x+1\right|$
First restrict $x$ to  $0<\left|x+1\right|<\frac{1}{4}$ $\Rightarrow$ initial choice of $\delta=\frac{1}{4}$
$\left| 2x+1 \right|$ = $\left|2(x+1)-1\right|$

$\le$ $\left|2(x+1)\right|+\left|-1\right|$ = $2\left|x+1\right|+1$ $> 1$

Thus if $\left|x+1\right|<\frac{1}{4}$ , then


$\left|\frac{x}{2x+1}-1\right|=\frac{1}{\left|2x+1\right|}.\left|x+1\right|$
    $<1.\left|x+1\right|$.


Therefore, $\delta = \min\{\frac{1}{4},\varepsilon\}$
$0<\left|x+1\right|<\delta$ $\Rightarrow$ $\left|\frac{x}{2x+1}-1\right|  < 1\cdot\left|x+1\right| < 1\cdot\varepsilon = \varepsilon$
Thus, the limit is 1.

You shouldn't change direction of your inequalities in a chain--for example, $$|2(x+1)-1|\le 2|x+1|+1>1$$ doesn't allow you to conclude that $|2(x+1)-1|>1,$ as transitivity breaks down when you switch directions.
Instead, we can use triangle inequality (why?) to say $$|2(x+1)-1|\ge1-2|x+1|,$$ so whenever $|x+1|<\frac14,$ we will have $$|2(x+1)-1|\ge1-2|x+1|>\frac12,$$ and so $$\left|\frac{x}{2x+1}-1\right|=\frac1{|2(x+1)-1|}|x+1|<2|x+1|.$$ A good choice would then be $$\delta=\min\left\{\frac14,\frac{\epsilon}2\right\}.$$
