Use of definitions to prove limits

**CrazyCat87** · February 7th 2010, 08:16 AM

How would you prove the following limit using the epsilon-delta definition of a limit,

$\lim_{x\to1} \frac{x}{1+x} = \frac{1}{2}$

**VonNemo19** · February 7th 2010, 08:25 AM

Originally Posted by CrazyCat87

How would you prove the following limit using the epsilon-delta definition of a limit,

$\lim_{x\to1} \frac{x}{1+x} = \frac{1}{2}$

Where are you stuck?

**CrazyCat87** · February 7th 2010, 05:30 PM

Originally Posted by VonNemo19

Where are you stuck?

Sorry, here is what I have so far
$|\frac{x}{1+x} - \frac{1}{2}| = |\frac{2x-x-1}{2x+2}| = |\frac{x-1}{2x+2}| = |\frac{1}{2x+2}||x-1|$

And we know $|x-1| \leq \delta$ .....

That's all I have right now. I can't figure out where to go from there

**Chris L T521** · February 7th 2010, 05:41 PM

Originally Posted by CrazyCat87

Sorry, here is what I have so far
$|\frac{x}{1+x} - \frac{1}{2}| = |\frac{2x-x-1}{2x+2}| = |\frac{x-1}{2x+2}| = |\frac{1}{2x+2}||x-1|$

And we know $|x-1| \leq \delta$ .....

That's all I have right now. I can't figure out where to go from there

Fix $\delta =1$ so we can find an upper bound for $\frac{1}{\left|x+1\right|}$ :

$\left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<3\implies\frac{1}{3}<\frac{1}{x+1}<1$ .

Therefore $\frac{1}{\left|2x+2\right|}\left|x-1\right|=\frac{1}{2\left|x+1\right|}\left|x-1\right|<\frac{1}{2}\left|x-1\right|<\varepsilon$ . Therefore, we pick $\delta=\min\{2\varepsilon,1\}$

Does this make sense?

**CrazyCat87** · February 7th 2010, 06:18 PM

Originally Posted by Chris L T521

Fix $\delta =1$ so we can find an upper bound for $\frac{1}{\left|x+1\right|}$ :

$\left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<2\implies\frac{1}{2}<\frac{1}{x+1}<1$ .

Therefore $\frac{1}{\left|2x+2\right|}\left|x-1\right|=\frac{1}{2\left|x+1\right|}\left|x-1\right|<\frac{1}{2}\left|x-1\right|<\varepsilon$ . Therefore, we pick $\delta=\min\{2\varepsilon,1\}$

Does this make sense?

Ok yea the only thing that never makes sense to me is this step
$\left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<2$

how do you go from $-1<x-1<1$ to $1<x+1<2$

**Chris L T521** · February 7th 2010, 06:21 PM

Originally Posted by CrazyCat87

Ok yea the only thing that never makes sense to me is this step
$\left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<2$

how do you go from $-1<x-1<1$ to $1<x+1<2$

Whoops...that should be $1<x+1<3$ , but even then, the upper bound on $\frac{1}{\left|x+1\right|}$ remains the same.

Sorry about that!

**CrazyCat87** · February 7th 2010, 07:12 PM

Right, I just wanted to understand what was going on. It all makes sense now, thank you so much for the help

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