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So I have something like this: $\lim_{n \rightarrow \infty} (-2 n+\sqrt{2+5 n+4 n^2})$

On the lesson, it was solved by multiplying by the conjugate and therefore arriving at the polynomial/polynomial form. I understand that, but for my current knowledge of limitis, it stands in contradiction to the properties of limits which state that $\lim_{n \rightarrow \infty}(a_n+b_n)=\lim_{n \rightarrow \infty}a_n+\lim_{n \rightarrow \infty}b_n$ and that $\lim_{n \rightarrow \infty}\sqrt{a_n}^k = \sqrt{\lim_{n \rightarrow \infty}a_n}^k$.

Why can't we use these two here? If we could, wouldn't it become $\lim_{n \rightarrow \infty}-2n=-\infty$ and $\lim_{n \rightarrow \infty}\sqrt{2+5 n+4 n^2}=\sqrt{\lim_{n \rightarrow \infty}2+5 n+4 n^2}=\infty$ and then $\infty-\infty=0$? Why would such reasoning be wrong?

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    I see, thank you a lot!2012-11-04

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The rule for distributing the limit over addition (or multiplication, etc.) is: if $(a_n)$ and $(b_n)$ both converge (to a finite limit) then $(a_n+b_n)$ converges (to a finite limit) and $\displaystyle \lim_{n \to \infty}(a_n + b_n) = \lim_{n \to \infty}(a_n) + \lim_{n \to \infty}(b_n)$.

The reason why this doesn't apply here is because both the sequences $(-2n)$ and $(\sqrt{2+5n+4n^2})$ diverge.

If you find yourself writing things like '$\infty - \infty = 0$' then you know you've gone wrong somewhere; it is not valid reasoning.

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    Heh, yeah, sure :) Thanks.2012-11-04
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$\infty - \infty $ is one of so called undefinite limits and the value depends on how mutch hard diverges the given sequences in this case we have $\lim_{n\to \infty}(-2n+\sqrt{2+5n+4n^2})$

$\lim_{n\to \infty}(-2n+\sqrt{2+5n+4n^2}) \frac{2n+\sqrt{2+5n+4n^2}}{2n+\sqrt{2+5n+4n^2}}=$

$\lim_{n\to \infty} \frac{4n^2-(2+5n+4n^2)}{2n+\sqrt{2+5n+4n^2}}=\lim_{n\to \infty} \frac{-(2+5n)}{2n+\sqrt{2+5n+4n^2}}=$

$=\lim_{n\to \infty} \frac{-(\frac{2}{n}+5)}{2+\sqrt{\frac{2}{n^2}+\frac{5}{n}+4}}=\frac{-5}{2+2}=-\frac{5}{4}$