# Could you help me understand a simple limit problem with conjugate?

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?

-
Arithmetic of limits applies only when the limits exists and they are finite (and in case of the quotient, one also has to check the denominator limit's not zero) – DonAntonio Nov 4 '12 at 17:39
I see, thank you a lot! – Straightfw Nov 4 '12 at 17:40

## 2 Answers

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.

-
Thank you. But how can I check if the things I'd want to apply such formulas on converge, which allows me to, or diverge, which makes me unable to? – Straightfw Nov 4 '12 at 17:37
Well it's as simple as: if you have the sum of two sequences, and either sequence diverges, then you can't say that the limit of the sum is the sum of the limits. Here it fails because $-2n \to -\infty$, and hence diverges. However you can say $\lim(\frac{1}{n} + 3n\sin \frac{1}{n}) = \lim \frac{1}{n} + \lim(3n \sin \frac{1}{n})$, since both terms converge. – Clive Newstead Nov 4 '12 at 17:42
Oh, I see! So if I had some example where I'm not sure, I could use the Cauchy's or d'Alambert's principle to check if they converge and if any of them doesn't, I have to use other way around than these limit-arithmetic things, is that right? – Straightfw Nov 4 '12 at 17:44
Uh... maybe? I can't quite tell what you mean. Basically: if you want to apply a result which depends on a certain condition being satisfied, then you have to check that the condition is satisfied before you apply the result. Here you tried to apply a result to a situation where the condition failed, and that condition was 'both sequences in the sum converge'. – Clive Newstead Nov 4 '12 at 17:46
Heh, yeah, sure :) Thanks. – Straightfw Nov 4 '12 at 17:46

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

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

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

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

-