Proving and Finding a limit I need to find the following limit and prove using the definition of limits. 
$$\lim_{x\to1} {x \over x+1} = \frac 1 2$$.
Following the definition: 
$$\forall \epsilon \exists \delta : \lvert x - c \rvert < \delta \Rightarrow \lvert F(x) - L \rvert < \epsilon$$
$$\left\lvert \frac{x}{x+1} - \frac{1}{2} \right\rvert < \epsilon = \left\lvert 2x-x-1 \over 2x+2 \right\rvert = \left\lvert x-1 \over 2x+2 \right\rvert < \epsilon$$
I have trouble around here. I don't know how to reach $\left\lvert x - c \right\rvert < \delta$
I tried:
$$ \frac{x-1}{2x+2} < \frac{x}{2x} = \frac{1}{2} < \epsilon$$
But something about that doesn't seem right to me. Can I get any hints?
 A: Hint:
Take $\delta=min(\epsilon,1)$, then $\forall |x-1|<\delta\le 1$, $|x+1|>1$, hence ...
The idea is to select $\delta$ such that $x$ is closer to $1$ than to $-1$ (the denominator is $2|x+1|$), so that you have a lower bound for the denominator.
A: Showing the versatility of this problem. 
$$\begin{align}|x - 1| < \delta \leq \frac{1}{2} &\Leftrightarrow -\frac{1}{2} < x - 1 < \frac{1}{2} \Leftrightarrow \frac{1}{2} < x < \frac{3}{2} \\ &\Leftrightarrow \frac{3}{2} < x + 1 < \frac{5}{2} \Rightarrow \frac{2}{5} < \frac{1}{x+1} < \frac{2}{3} \end{align}$$
In particular $\Big|\frac{1}{x+1}\Big| < \frac{2}{3}$. 
Then as you have already reached
$$\left\lvert \frac{x}{x+1} - \frac{1}{2} \right\rvert = \left\lvert 2x-x-1 \over 2x+2 \right\rvert = \left\lvert x-1 \over 2x+2 \right\rvert = \frac{1}{2}\left\lvert x-1 \over x+1 \right\rvert  < \frac{\delta}{3}$$
Now take $\delta = \min \lbrace \epsilon , \frac{1}{2}\rbrace$. And you will have for every $\epsilon > 0$ given, there exits $\delta = \min \lbrace \epsilon, \frac{1}{2}\rbrace$ such that 
$$|x - 1| < \delta \Rightarrow \Bigg|\frac{x}{x + 1} - \frac{1}{2}\Bigg| < \epsilon$$
