# 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?

• You've missed all your dollar signs. I'm not able to edit it. Nov 21 '14 at 9:48
• This is a similar question
– John
Nov 21 '14 at 9:57

## 2 Answers

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.

• Sorry, but you could you elaborate? I'm having trouble wrapping my head around it... Edit: Nevermind! I got it! Thanks! Nov 21 '14 at 10:10
• NB I fixed a minor grammar issue / typo. Nov 21 '14 at 10:11

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

$$|x - 1| < \delta \Rightarrow \Bigg|\frac{x}{x + 1} - \frac{1}{2}\Bigg| < \epsilon$$