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f(n+1)=1/2{f(n)+9/f(n)},n belongs to natural nos. and f(n)>0 for all n belonging to natural nos. then limit n tends to infinity equals to what
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if the limit exists, then f(n+1) approaches f(n). call this value x then x= (1/2) (x+9/x) 2x= x+9/x x=9/x x^2= 9 x= sqrt(9)
thank you
f(n)>0 for all n so it looks like x= +3 is the limit
the answer is correct
@phi-can you help me with a question from complex numbers
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http://openstudy.com/study#/updates/4f621417e4b079c5c6312ef5 this is the link to the question
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