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

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

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- Feb 7th 2010, 08:16 AMCrazyCat87Use of definitions to prove limits
How would you prove the following limit using the epsilon-delta definition of a limit,

$\displaystyle \lim_{x\to1} \frac{x}{1+x} = \frac{1}{2}$ - Feb 7th 2010, 08:25 AMVonNemo19
- Feb 7th 2010, 05:30 PMCrazyCat87
Sorry, here is what I have so far

$\displaystyle |\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 $\displaystyle |x-1| \leq \delta $ .....

That's all I have right now. I can't figure out where to go from there - Feb 7th 2010, 05:41 PMChris L T521
Fix $\displaystyle \delta =1 $ so we can find an upper bound for $\displaystyle \frac{1}{\left|x+1\right|}$:

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

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

Does this make sense? - Feb 7th 2010, 06:18 PMCrazyCat87
- Feb 7th 2010, 06:21 PMChris L T521
- Feb 7th 2010, 07:12 PMCrazyCat87
Right, I just wanted to understand what was going on. It all makes sense now, thank you so much for the help