PDA

View Full Version : Limit Calculation



ku1005
04-15-2007, 02:21 AM
Using properties of limits show that thereexists an N in the set of positive integers such that

n>N => ((n+1)^2)/2n^2 < 3/4

For this question, would someone be able to show me a more succinct approach, as I belive the process I used isnt very efficient, say I had to apply it to a different Q which was harder, or maybe Im wrong and the way I solved it is fine???...any suggestions great!!

I simple took the limit of the sequence as it went to infinity, ie divide all by n^2

therfore you obtain (1 + 2/n + 1/n^2) / 2

which as -> inifnity obviously = 1/2 which is less then 3/4

therfore, I SUB each number (this this bit I dont like- i belive ther is usually a way you can solve for this number) and found that once n> 4, then the seq is always less then 3/4

hence N = 4

cheers for any help suggestions as always!!!

pka
04-15-2007, 08:23 AM
Why not just solve the inequality?
\L
\begin{array}{l}
\frac{{\left( {n + 1} \right)^2 }}{{2n^2 }} < \frac{3}{4} \\
4\left( {n + 1} \right)^2 < 6n^2 \\
\end{array}

ku1005
04-16-2007, 12:19 AM
i see...good point....you will obtain 2 results....and in knowing must be positive intergers the n>4 comes about....thanks for that..much appreciated!