<s>[INST] I can't understand the result of this limit $$ \lim_{n\to\infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}
$$
I think the answer is $-\infty$ because I take the highest degree coefficients $$n^4  (numerator)$$ and $$-n ^ 3(denominator)$$
and I was thinking $+/- = -$, but the actual answer is $+\infty$, this is probably a dumb question, but I didn't understand why is $+\infty$ instead of $-\infty$.
 [/INST] \begin{align} \lim_{n\to\infty} \left(\frac{n^4-n^3+1}{-n^3+n^2+\sqrt{n}}\right) \\
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=\lim_{n\to\infty} \left(n \ . \ \frac{n^3-n^2+\frac1n}{-n^3 + n^2 + \sqrt{n}}\right) \\
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=\lim_{n\to\infty} \left(n \ . \ \left(-\ \frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
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=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
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=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}} \ .\ \frac{ \frac{1}{n^3} }{ \frac{1}{n^3} }\right)\right) \\
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=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{1-\frac{1}{n}+\frac{1}{n^4}}{1-\frac{1}{n} -\frac{\sqrt{n}}{n^3}} \ \right)\right) \\
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=-\ \lim_{n\to\infty} \left(n\right) \ .\ 1 \\
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=-\infty.
\end{align}
</s>