# Use of definitions to prove limits

• Feb 7th 2010, 08:16 AM
CrazyCat87
Use 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 AM
VonNemo19
Quote:

Originally Posted by CrazyCat87
How would you prove the following limit using the epsilon-delta definition of a limit,

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

Where are you stuck?
• Feb 7th 2010, 05:30 PM
CrazyCat87
Quote:

Originally Posted by VonNemo19
Where are you stuck?

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 PM
Chris L T521
Quote:

Originally Posted by CrazyCat87
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

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 PM
CrazyCat87
Quote:

Originally Posted by Chris L T521
Fix $\displaystyle \delta =1$ so we can find an upper bound for $\displaystyle \frac{1}{\left|x+1\right|}$:

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

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?

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

how do you go from $\displaystyle -1<x-1<1$ to $\displaystyle 1<x+1<2$
• Feb 7th 2010, 06:21 PM
Chris L T521
Quote:

Originally Posted by CrazyCat87
Ok yea the only thing that never makes sense to me is this step
$\displaystyle \left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<2$

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

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