limit of sqrs

math_Lover

New member
Joined
Dec 8, 2020
Messages
5
What is the limit of [MATH]\sqrt6,\sqrt{6+\sqrt6},\sqrt{6+\sqrt{6+\sqrt6}},\cdots[/MATH]I would appreciate an answer that includes both a way and a final answer.

Thanks :)
 
What is the limit of [MATH]\sqrt6,\sqrt{6+\sqrt6},\sqrt{6+\sqrt{6+\sqrt6}},\cdots[/MATH]I would appreciate an answer that includes both a way and a final answer.

Thanks :)
a1 = √6, a2 =√(6+a1), a3 =√(6+a2) ...... an = ?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
consider the sequence of terms produced by the rule.

[MATH]a_{n+1} = \sqrt{6 + a_n},~a_1=\sqrt{6}[/MATH]
Now in the limit what can you say about [MATH]a_n[/MATH] and [MATH]a_{n+1}[/MATH] ?
 
What kind of determination is "something like". Please do better than that.

If [math]x = \sqrt{6+\sqrt{6+\sqrt{6+...}}}[/math],
Then [math]x = \sqrt{6+x}[/math]
Please show YOUR work in the future.
 
What is the limit of [MATH]\sqrt6,\sqrt{6+\sqrt6},\sqrt{6+\sqrt{6+\sqrt6}},\cdots[/MATH]I would appreciate an answer that includes both a way and a final answer.
Suppose \(\ell=\sqrt6,\sqrt{6+\sqrt6},\sqrt{6+\sqrt{6+\sqrt6}},\cdots\) then \(\ell=\sqrt{6+\ell~~}\)
So that \(\ell^2-\ell+6=0\) or \((\ell-3)(\ell+2)=0\) Which of values is the answer and explain why.
 
Last edited:
Top