Thread: Use of definitions to prove limits

1. Use of definitions to prove limits

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

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

2. 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?

3. 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

4. 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.

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?

5. 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.

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

how do you go from $-1 to $1

6. 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

how do you go from $-1 to $1
Whoops...that should be $1, but even then, the upper bound on $\frac{1}{\left|x+1\right|}$ remains the same.