Proving that a certain function approaches a certain limit at a certain point using epsilon-delta method (Proof Check) I have to prove that, 
$\lim_{x\to 1}$ $\frac{x}{x+1}$ = $\frac{1}{2}$
using delta-epsilon method.
My answer: Let $\epsilon$ > $0$ and let $\delta$ = min( 1 , 2$\epsilon$)
Then,  $\left|\frac{x}{x+1} - \frac{1}{2}\right|$ = $\left|x - 1\right|\frac{1}{2}\left|\frac{1}{x+1}\right|$ 
Since $\left|x-1\right| < 1$ implies that $0 < x < 2$ 
which, in turn implies that $\frac{1}{3} < \frac{1}{x+1} < 1$
So the result follows, $\left|x - 1\right|\frac{1}{2}\left|\frac{1}{x+1}\right| < \epsilon$ 

Depending on where you are in your studies/class, more justification around some of these conclusions could be helpful for the sake of clarity. For example, writing:
$$\Big|\frac{x}{x+1} - \frac{1}{2}\Big| = \Big|\frac{2x}{2(x+1)} - \frac{x+1}{2(x+1)}\Big| = \Big|\frac{x-1}{2(x+1)}\Big| = \frac{1}{2}\Big|x-1\Big|\Big|\frac{1}{x+1}\Big|$$
Or, looking to the next line, that: $|x - 1| < 1$ means $-1 < x - 1 < 1$; so, we have $0<x<2$.
Similarly, this final inequality is equivalent to $1 < x+1 < 3$, whence 
$$\frac{1}{3} < \frac{1}{x+1} < 1$$
as claimed. Acceptable proof writing is, to a nontrivial extent, a function of the reader.
