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mathivh
 one year ago
Limit (to infinity) question
mathivh
 one year ago
Limit (to infinity) question

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mathivh
 one year ago
Best ResponseYou've already chosen the best response.1So when you have: \[\lim_{x \rightarrow \infty}(\frac{ n^2 }{ n^2 +2n+1})\] you can divide this by \[n^2\] so you would get \[\lim_{x \rightarrow \infty}(\frac{ n^2/n^2 }{ n^2/n^2+ 2n/n^2 + 1/n^2 }) = (\frac{1}{1+0+0}) = 1\] But why can't you divide \[\lim_{x \rightarrow \infty}(n+1)\] by n, so that you would get:\[\lim_{x \rightarrow \infty}(n/n +1/n)= 1 + 0 = 1\]

mathivh
 one year ago
Best ResponseYou've already chosen the best response.1Obviously I meant \[\lim_{n \rightarrow \infty}\] instead of \[\lim_{x \rightarrow \infty}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.5You CAN'T just divide by n. If you want to do that, you MUST multiply by n also. Like, \(a =a *\dfrac{n}{n}\) to not change the original problem.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.5In previous case, you divide by n BOTH numerator and denominator; it doesn't change the original problem, right? \(\dfrac{a}{b}=\dfrac{a/c}{b/c}\)

mathivh
 one year ago
Best ResponseYou've already chosen the best response.1Aah ok I understand now, pretty silly of me to have overseen that. Thanks for the help @Loser66 !
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