- #1
FaroukYasser
- 62
- 3
Edit: the red \E is the epsilon sign
1. Homework Statement
##Given\quad { x }_{ n }=\sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \\ Prove\quad that:\quad \lim _{ n\rightarrow \infty }{ { x }_{ n } } =0##
##For\quad any\quad arbitrary\quad \E >0,\quad there\quad exists\quad an\quad N>0\quad such\quad that\\ if\quad n>N\quad \Longrightarrow \quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| <\E ##
##\left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| \quad \le \quad \sqrt { n+1 } +\sqrt { n+2 } +2\sqrt { n+3 } \\ \\ \\ <\quad \sqrt { n+3 } +\sqrt { n+3 } +2\sqrt { n+3 } \quad =\quad 4\sqrt { n+3 } \quad \\ \\ <\quad \E \\ \\ Taking\quad N\quad =\quad -3\quad +\quad \left( \frac { \E }{ 4 } \right) ^{ 2 }\\ if\quad n>N\quad \Longrightarrow \quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| <\quad \E \\ thus\quad by\quad the\quad definition\quad of\quad a\quad limit,\quad the\quad limit\quad is\quad 0.##
[/B]
However, whenever I take any epsilon, say 0.1, I get N=-2.999375, so if I choose n = 1 for example, and I sub it into the inequality, I get 0.85337... < 0.1 which is obviously incorrect. Any idea what part is wrong?
1. Homework Statement
##Given\quad { x }_{ n }=\sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \\ Prove\quad that:\quad \lim _{ n\rightarrow \infty }{ { x }_{ n } } =0##
Homework Equations
##For\quad any\quad arbitrary\quad \E >0,\quad there\quad exists\quad an\quad N>0\quad such\quad that\\ if\quad n>N\quad \Longrightarrow \quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| <\E ##
The Attempt at a Solution
##\left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| \quad \le \quad \sqrt { n+1 } +\sqrt { n+2 } +2\sqrt { n+3 } \\ \\ \\ <\quad \sqrt { n+3 } +\sqrt { n+3 } +2\sqrt { n+3 } \quad =\quad 4\sqrt { n+3 } \quad \\ \\ <\quad \E \\ \\ Taking\quad N\quad =\quad -3\quad +\quad \left( \frac { \E }{ 4 } \right) ^{ 2 }\\ if\quad n>N\quad \Longrightarrow \quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| <\quad \E \\ thus\quad by\quad the\quad definition\quad of\quad a\quad limit,\quad the\quad limit\quad is\quad 0.##
[/B]
However, whenever I take any epsilon, say 0.1, I get N=-2.999375, so if I choose n = 1 for example, and I sub it into the inequality, I get 0.85337... < 0.1 which is obviously incorrect. Any idea what part is wrong?
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