Limit (to infinity) question

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Limit (to infinity) question

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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So 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:\[\lim_{x \rightarrow \infty}(n/n +1/n)= 1 + 0 = 1\]
Obviously I meant \[\lim_{n \rightarrow \infty}\] instead of \[\lim_{x \rightarrow \infty}\]
You 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.

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In previous case, you divide by n BOTH numerator and denominator; it doesn't change the original problem, right? \(\dfrac{a}{b}=\dfrac{a/c}{b/c}\)
Aah ok I understand now, pretty silly of me to have overseen that. Thanks for the help @Loser66 !
np

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