Semantic in limits calculation. The best way to explain this question is by example.
Calculate: $\lim_{n\to\infty}\frac{(-3)^{n+1}-4^n+5}{3^{n+2}+2^n-5}$.
So, 
$
\begin{aligned}
\lim_{n\to\infty}\frac{(-3)^{n+1}-4^n+5}{3^{n+2}+2^n-5}=
\lim_{n\to\infty}\frac{\frac{(-3)^{n+1}}{4^n}-\frac{4^n+5}{4^n}}{\frac{3^{n+2}}{4^n}+\frac{2^n-5}{4^n}}="\frac{0-1+0}{0+0-0}"="-\frac{1}{0}"=-\infty.
\end{aligned}
$
But,
$
\begin{aligned}\\
\lim_{n\to\infty}\frac{(-3)^{n+1}-4^n+5}{3^{n+2}+2^n-5}=
\lim_{n\to\infty}\frac{\frac{(-3)^{n+1}}{-4^n}-\frac{4^n+5}{-4^n}}{\frac{3^{n+2}}{-4^n}+\frac{2^n-5}{-4^n}}="\frac{0+1+0}{-0-0+0}"="\frac{1}{0}"=\infty.
\end{aligned}
$
The sign of the zeros in the denominator is unimportant because $-0=0=+0$. So i'm wondering what is the explanation of this. Thank you.
 A: Probably the best answer is: while $\frac01$ and $\frac{-0}1$ are the same, $\frac10$ and $\frac1{-0}$ are not the same—for reasons that have to do with why "$=$" is not the same as $=$.
Consider the simpler example $\lim_{n\to\infty} \frac{-n}{1}$.
The two expressions analogous to your calculations are
\begin{align*}
\lim_{n\to\infty} \frac{-n}{1} = \lim_{n\to\infty} \frac{\frac{-n}n}{\frac1n} &"=" \frac{-1}0 "=" -\infty \\
\lim_{n\to\infty} \frac{-n}{1} = \lim_{n\to\infty} \frac{\frac{-n}{-n}}{\frac1{-n}} &"=" \frac{1}{-0} "=" -\infty \\
\text{but not} \qquad \cdots &"=" \frac{1}{-0} "=" \frac{1}{0} "=" \infty.
\end{align*}
If you like, take this as additional reason to believe that $\infty$ is not a number and that dividing by $0$ really isn't allowed.
The quotes around the "$=$" symbols are emphasizing that we are using these symbols (and $\frac10$, etc.) as a shorthand for careful reasoning about infinite limits—but the reasoning is what's rigorous, not some pattern we might be tempted to deduce from the shorthand symbols themselves.
In your example, can you verify that the numerator of the original expression is negative when $n$ is large, while the denominator is positive when $n$ is large? Does that help you evaluate the limit of their ratio?
A: In the second example, consider the value of your denominator as it approaches zero.
Personally, I would prefer dividing by a factor that gives you a constant in your denominator (which would you pick?). The answer should be the same. 
