$\epsilon - \delta$ proof of $\lim _{x \rightarrow 0} \frac{2 x+3}{x+1}=3$ A hint is given: $|x| < 1/2 \implies |x+1| > 1/2$.
I have been advised to only use the general method. which is starting from 
$$\left|\frac{2 x+3}{x+1}-3\right|$$
and ending with something with $|x-0|$. 
I have come this far 
$$
\left|\frac{2 x+3}{x+1}-3\right|=\left|\frac{-x}{x+1}\right|=\frac{|x|}{|x+1|}
$$
I think that I have to use the hint now, but I can't figure out how and with what purpose.
 A: The way $\delta-\epsilon$ proofs is that for ever $\epsilon >0$ there is a $\delta >0$ so that $|x-a| < \delta \implies |f(x)-L| < \epsilon$.
We have two possible strategies. Either start at the end and work back:
$|f(x) - L| <\epsilon \Leftarrow$
....
$|x - a| < h(\epsilon) = \delta$
Or
Start at the beginning and work forward.
$|x-a| < \delta \implies$
....
$|f(x) - L| < j(\delta) \le \epsilon$ and we work backwards to figure out what $\delta$ is in terms of $\epsilon$.
...
In this case the hint is:  If $|x| < \frac 12$ then $|x+1| = x+1 >\frac 12$ and that $\frac 1{|x+1|} < 2$.
So if we choose our $\delta \le \frac 12$ then we can replace any occurance of $\frac M{|x+1|}$ with $\frac M{|x+1|}  < \frac M2$.
That makes this work, whether front to back  or back to front short work:
If $|x| < \delta \le \frac 12$ then
$2{|x|}2 < 2\delta $
$\frac {|x|}{|x+1|} < \frac {|x|}{\frac 12}=2|x| \le 2 \delta $ and
$|\frac {2x+1}{x+1} -3|=|\frac {-x}{x+1}| = \frac {|x|}{|x+1|}<2{|x|} \le 2 \delta $.
So as long as we choose $\delta$ so that $2\delta  \le \epsilon$ and $\delta < \frac 12$, or in other words if $\delta \le \min(\frac 12, \frac\epsilon 2)$ we will get our result.
Or going the other way:
$|\frac {2x+1}{x+1} - 3| = |\frac {-x}{x+1}|=\frac {|x|}{|x+1|}< \epsilon$ 
Is implied by $\frac{|x|}{|x+1|} < 2{|x|} < \epsilon$ if $|x| < \frac 12$
Which would follow if $|x| < \frac{\epsilon}2$ and $|x| < \frac 12$.  
So if $\delta = \min (\frac\epsilon 2, \frac 12)$ we'll be fine.
A: It follows from that hint that$$\lvert x\vert<\frac12\implies\left\lvert\frac{2x+3}{x+1}-3\right\rvert<2\lvert x\rvert.$$So, take $\delta=\min\left\{\frac\varepsilon2,\frac12\right\}$. Then, if $\lvert x\rvert<\delta$, then$$\left\lvert\frac{2x+3}{x+1}-3\right\rvert<2\lvert x\rvert.\tag1$$And now, since $\lvert x\rvert<\frac\varepsilon2$, you deduce from $(1)$ that$$\left\lvert\frac{2x+3}{x+1}-3\right\rvert<\varepsilon.$$
