<s>[INST] Find $\lim_{n\to \infty}\frac{-7n^5-6n^4+122n^3-15n^2+7n-2}{21n^5+12n^3-501}$ I am stuck with this question, I do not know what I'm doing wrong. I am dividing everything by $n^5$ and getting end result as $\frac{-7}{21}$ but the result in book answer says -$\frac{1}{3}$. Can someone advice? My calculation:
$${\displaystyle\lim_{n\to \infty}}\frac{-7n^5-6n^4+122n^3-15n^2+7n-2}{21n^5+12n^3-501}=\frac{n^5(-7-\frac{6n^4}{n^5}+\frac{122n^3}{n^5}-\frac{15n^2}{n^5}+\frac{7n}{n^5}-\frac{2}{n^5})}{n^5(21+\frac{12n^3}{n^5}-\frac{501}{n^5})}$$
So obviously here $n^5$ cancels out, $\frac{-7}{21}$ stays and all other values go to $0$.
 [/INST] You are correct, infact we have
$$\frac{-7}{21} = \frac{-7}{7 \times 3}=\frac{-1}{3}$$
</s>