What is wrong with this limit $\lim_{x \rightarrow \infty} \frac{2x^4-1}{-4x^5+x^2}$? The arrow indicates an elimination due to numbers becoming negligible at high values of x. 
The correct answer is 0, but I got infinity and I'm not sure where my reasoning went wrong.
$$\lim_{x \rightarrow \infty} \frac{2x^4-1}{-4x^5+x^2}=\lim_{x \rightarrow \infty} \frac{x^4(2-\frac{1}{x^4})}{x^5(-4+\frac{x^2}{x^5})}=\lim_{x \rightarrow \infty} \frac{x^4(2)}{x^5(\frac{1}{x^3})}=\lim_{x \rightarrow \infty} \frac{2x^4}{x^2}=\lim_{x \rightarrow \infty} 2x^2 = \infty$$
Thanks!
 A: Here is a solution.
$$\lim_{x\to \infty}\frac{2x^4-1}{-4x^5+x^2} = \lim_{x \to \infty}\frac{\frac{2}{x}-\frac{1}{x^5}}{-4+\frac{1}{x^3}} = \frac{0-0}{-4+0}=0$$
I would say that the error is that you can't just take the result of the limit for some parts of the expression and some not, which is what I believe you did in the second equality.
A: Since I am sure you understood the answers to your problem, let us try to consider the general problem of finding $$\lim_{x \rightarrow \infty} \frac{P_m(x)}{Q_n(x)}$$ where $P_m(x)$ and $Q_n(x)$ are polynomials of degrees $m$ and $n$ that is to say $$P_m(x)=\sum _{i=0}^m p_i x^i=p_0+p_1x+p_2x^2+...+p_mx^m$$  $$Q_n(x)=\sum _{i=0}^n q_i x^i=q_0+q_1x+q_2x^2+...+q_nx^n$$in which coefficients $p_m$ and $q_n$ are not zero. You can factor $x^m$ in the numerator and $x^n$ in the denominator and then write $$P_m(x)=x^m  \sum _{i=0}^m p_i x^{i-m}$$ $$Q_n(x)=x^n  \sum _{i=0}^n q_i x^{i-m}$$ In the summations, except for the last terms which are $p_m$ and $q_n$ only negative powers of $x$ appear and, since $x$ goes to $\infty$, they are negligible. So, for large values of $x$, $$\frac{P_m(x)}{Q_n(x)} \simeq \frac {p_m x^m}{q_n x^n}=\frac {p_m }{q_n} x^{m-n}$$ 
As a consequence, if $m>n$ the limit is $\infty$, if $m=n$ the limit is $\frac {p_m }{q_n}$ and if $m<n$ the limit is $0$.
A: First arrow is incorrect since $-4+x^{-3} \to -4$ as $x\to\infty$.
A: Your mistake is not that you did not evaluate the numerator and denominator at the same time or in the same way.  Your mistake is strictly a computational one: $$\lim_{x \rightarrow \infty} \frac{x^4(2-\frac{1}{x^4})}{x^5(-4+\frac{x^2}{x^5})}=\lim_{x \rightarrow \infty} \frac{x^4(2)}{x^5(\frac{1}{x^3})}$$ is incorrect.  The RHS must instead read $$\lim_{x \rightarrow \infty} \frac{x^4(2)}{x^5(-4 + \frac{1}{x^3})},$$ and this is the source of the error.  This is because as $x \to \infty$, $x^5(-4 + x^{-3})$ asymptotically behaves as $-4x^5$, not $x^2$, and it is this faster growth of the denominator compared to the numerator that makes the limit zero.
