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
$$
\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} = \infty
$$
 A: Continuing from where you've reached, you can conclude that $$\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4}$$
where as $x \to \infty$, we have $\frac{8}{x} \to 0$. The same goes for $\frac{1}{x^2} \to 0$, and $\frac{7}{x^4} \to 0$. We also have $\frac{11x}{x^4} \to 0$, so you can rewrite the above approximately for large $x$ as $$\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} \approx \frac{0-0}{0+0-4} = 0$$

Alternatively, (this isn't a technique you are likely to understand just yet and I wouldn't recommend using it for the moment, but I am including it any way just for an alternative way), we could use L'Hôpital's rule. Since we have $$\lim_{x\to \infty} \frac {8x^3-x^2}{7+11x-4x^4} = \frac{\infty}{\infty}$$ an indeterminate form L'Hôpital once to get $$\lim_{x\to \infty} \frac {8x^3-x^2}{7+11x-4x^4} = \lim_{x\to \infty}\frac{24x^2 - 2x}{11 - 16x^3}$$
which is still an indeterminate form, apply L'Hôpital again to get $$\lim_{x\to \infty}\frac{24x^2 - 2x}{11 - 16x^3} = \lim_{x\to \infty}\frac{48x - 2}{- 48x^2}$$ still indeterminate, so a final application of L'Hôpital gives us: $$\lim_{x\to \infty}\frac{48x - 2}{- 48x^2} = \lim_{x\to \infty} -\frac{2}{x}$$ which tends to $0$ as $x \to \infty$.
A: 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 $n>m$, $n=m$, $n<m$.
A: Suppose that $n,m\in \mathbb{N}$:
$$f(x)=\dfrac{a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}}{b_{m}x^m+a_{m-1}x^{m-1}+\dots+b_{1}x+b_{0}}$$
We have three case:
Case $1$:
$$n< m\Longrightarrow  \lim_{x\rightarrow\infty}f(x)=0$$
Case $2$:
$$m<n\Longrightarrow \lim_{x\rightarrow\infty}f(x)=\infty$$
Case $3$:
$$n=m\Longrightarrow \lim_{x\rightarrow\infty}f(x)=\dfrac{a_{n}}{b_{m}}$$
