# Where is the mistake in this epsilon delta proof?

1. Sep 4, 2015

### FaroukYasser

• Missing template due to originally being posted in different forum.
The question asks to proof that the limit given in incorrect by contradiction. I tried working using the estimation method and the weird thing is that I completed the proof and found that the supposedly "incorrect" limit gave a correct answer although it was supposed to give me a contradiction of some sort, can someone help?

$Let\quad { X }_{ n }=\sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } ,\quad by\quad contradiction\\ show\quad that:\\ \lim _{ n\rightarrow \infty }{ \left[ \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right] } =\infty \quad is\quad not\quad true.\\ \\ My\quad answer:\\ \\ for\quad any\quad arbitrary\quad given\quad M>0,\quad there\quad exists\quad an\quad N.\\ such\quad that\quad if\quad n>N,\quad then\quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| >M\\ \\ \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| \quad >\quad 2\sqrt { n+3 } -\sqrt { n+1 } -\sqrt { n+2 } \\ >\quad -2\sqrt { n+3 } -\sqrt { n+1 } -\sqrt { n+2 } =\quad -\left( 2\sqrt { n+3 } +\sqrt { n+1 } +\sqrt { n+2 } \right) \\ >\quad -\left( 2\sqrt { n+3 } +\sqrt { n+3 } +\sqrt { n+3 } \right) =-4\sqrt { n+3 } \quad >\quad M\\ taking\quad N\quad =\quad -3\quad +\quad \frac { { M }^{ 2 } }{ 16 } \\ if\quad n>N\quad \Longrightarrow \quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| >M\\$

How can I have reached an N when the limit is incorrect?! Any help would be appreciated.
Also, sorry for my bad English.

2. Sep 4, 2015

### Orodruin

Staff Emeritus
You do realise that $-4\sqrt{n+3}$ is negative and therefore always smaller than $M$? There are several such mistakes.

3. Sep 5, 2015

### FaroukYasser

Ohh wow! Its 3 AM here so I do apologize for this.

Also, sorry for posting in the wrong forum.
Cheers :)

4. Sep 6, 2015

### Fredrik

Staff Emeritus
Did you mean $\lim_n X_n=+\infty$ or $\lim_n|X_n|=+\infty$?