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 $-\infty$.
 A: At the moment, we have
$$  \lim_{x \rightarrow \infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3} = \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}  $$
because the expression in the limit has no $x$s (and the limit is as $x$ goes somewhere).
Assuming what was intended was "$\lim_{n \rightarrow \infty}$", we compute, "factoring out the big term" in the numerator and denominator,
\begin{align*}
\lim_{n \rightarrow \infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3} 
    &= \lim_{n \rightarrow \infty} \frac{n^4}{n^3} \cdot \frac{1-n^{-1}+n^{-4}}{n^{-5/2} + n^{-1}-1}  \\
    &=  \lim_{n \rightarrow \infty} n \cdot \frac{1-0+0}{0 + 0-1}  \\
    &=  -1 \cdot \lim_{n \rightarrow \infty} n  \\
    &= -\infty  \text{.}  
\end{align*}
We can also guess at this in the graph of the limitand.

The expression in the limit does approach the simpler expression $-n$ as $n$ increases (after some excitement on $[0,2]$, which does not matter for the limit as $n \rightarrow \infty$ because $[0,2]$ is not near $\infty$).
A: First of all it should either be $ n \to \infty $ or you should replace $ n $ with $ x $. It is
$ \displaystyle\lim_{x \rightarrow\infty  }\frac{x^4-x^3+1}{\sqrt{x}+x^2-x^3} = \lim_{x \rightarrow\infty  }\frac{x^4}{-x^3} = \lim_{x \rightarrow\infty  } (-x) = -\infty $
A: \begin{align} \lim_{n\to\infty} \left(\frac{n^4-n^3+1}{-n^3+n^2+\sqrt{n}}\right) \\
\\
=\lim_{n\to\infty} \left(n \ . \ \frac{n^3-n^2+\frac1n}{-n^3 + n^2 + \sqrt{n}}\right) \\
\\
=\lim_{n\to\infty} \left(n \ . \ \left(-\ \frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
\\
=-\ \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) \\
\\
=-\ \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) \\
\\
=-\ \lim_{n\to\infty} \left(n\right) \ .\ 1 \\
\\
=-\infty.
\end{align}
