<s>[INST] 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!
 [/INST] 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.
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