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.
 A: To "discover" the proof, we typically work backwards. We "assume" ${|x+1|\over|2x+1|}<\epsilon$, and find what $\delta$ should be. Then we will have to rewrite the proof. 
Let's think about the numerator and denominator separately. We want $|x+1|$ small, and intuitively, we want $|2x+1|$ large. So let's keep $2x+1$ away from $0$ first. I think that's the $\delta<{1\over4}$ part. Then 
$$-{1\over 4}<x-(-1)<{1\over4},$$ 
so $-{5\over 4}<x<-{3\over4}$, so $-{5\over 2}<2x<-{3\over2}$ so $-{3\over 2}<2x+1<-{1\over2}$. So we use the fact that $|2x+1|>{1\over2}$. Thus, 
$${|x+1|\over|2x+1|}<2|x+1|<\epsilon.$$
Then we notice 
$$\delta<\min\bigg\{{\epsilon\over2},{1\over4}\bigg\}$$
Should do the trick. At this point, we've just "discovered" the proof, and $\delta(\epsilon)$, so now we proceed to write the logic out concisely. Fix $\epsilon$, blah blah blah
A: 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 $$\delta=\min\left\{\frac14,\frac{\epsilon}2\right\}.$$
A: Hint: Note that 
\begin{align}
\left| \frac{x}{2\cdot x +1} - 1 \right| < \epsilon 
\Longleftrightarrow
&
1-\epsilon< \frac{x}{2\cdot x +1}< 1+\epsilon
\\
\Longleftrightarrow
&
\left\{
\begin{array}
(2x+1)(1-\epsilon)<x  \\
\\
x< (2x+1)(1+\epsilon)
\end{array}
\right.
\end{align}
After algebric manipulations of two last inequalities you get
$$
\left| \frac{x}{2\cdot x +1} - 1 \right| < \epsilon 
\Longleftrightarrow
\left\{
\begin{array}
\;x-1<\frac{1+\epsilon}{1+2\epsilon}\\
\\
x-1< \frac{1-\epsilon}{1-2\epsilon}
\end{array}
\quad \mbox{ for } \epsilon <\frac{1}{2} 
\right.
$$
This suggests $\delta=\min\{\frac{1-\epsilon}{1-2\epsilon},\frac{1-\epsilon}{1-2\epsilon}\}$.
