<s>[INST] Calculate Infinite Limit I'm trying to calculate the limit and when I get to the last step I plug in infinity for $\frac 8x$ and that divided by -4 I get - infinity for my answer but the book says 0. Where did I go wrong?
$$
\frac {8x^3-x^2}{7+11x-4x^4}
$$
Divide everything by $x^4$
$$
\frac {\frac{8x^3}{x^4}-\frac{x^2}{x^4}}{\frac{7}{x^4}+\frac{11x}{x^4}-\frac{4x^4}{x^4}}
$$
Results
$$
\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} = \infty
$$
 [/INST] Hint
When you have expressions which are ratios of polynomials and that $x\to \infty$, the best is to factor the highest powers in numerator and denominator. So, in your case $$A=\frac {8x^3-x^2}{7+11x-4x^4}=\frac{x^3\big(8-\frac 1x\big)}{x^4\big(\frac 7{x^4}+\frac {11}{x^3}-4\big)}$$ So, since $x$ is large, we can "ignore" $\frac 1x$ (very small if compared to $8$), as well as $\frac 7{x^4}+\frac {11}{x^3}$ (very small if compared to $-4$). All of that arrives to $$A\approx \frac {8x^3}{-4x^4}=-\frac 2x$$ from which you can conclude.
If you apply to $$B=\frac{a_0+a_1x+a_2x^2+\cdots+a_nx^n}{b_0+b_1x+b_2x^2+\cdots+b_mx^m}$$ you could easily show that, for an infinite value of $x$, $$B\approx \frac{a_nx^n}{b_mx^m}=\frac{a_n}{b_m}x^{n-m}$$ Now, look what happens if $n>m$, $n=m$, $n<m$.
</s>