Why does this sequence not have a limit? $(C_n)_{n\in N}\ with \ c_n = (-1)^n\frac{n^3 + 2}{n^2+1}$ for all $n \in N$

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Why does this sequence not have a limit?

$(C_n)_{n\in N}\ with \ c_n = (-1)^n\frac{n^3 + 2}{n^2+1}$ for all $n \in N$

There sadly isn't any explanation for this particular sequence in my text book, it just says that it has no upper, or lower limit. I understand that the $(-1)^n$ means that the respective element is positive whenever n is even and negative whenever n is uneven, but couldn't it still have a minimum in the negative numbers and a maximum in the positive numbers?

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For $\frac {n^3 + 2}{n^2 + 1} = \frac {n^3 + n^2 - n^2 -1 + 3}{n^2 + 1} = n -1 + \frac 3{n^2 + 1}$

For any $N$ then for any even $n > |N| + 1$ then $c_n > |N|$. and for any odd $n > N + 1$ then $c_n < -|N|$.

So like the text says, there is no upper or lower limit.