Help verify $\lim_{x\to\infty} \frac{20x^2+6x+4}{7x-6-5x^2}$ Help verify this answer.  Determine the limit $$\lim_{x\to\infty} \frac{20x^2+6x+4}{7x-6-5x^2}$$
I say the answer is $-4$ because what I did was took the highest $x$'s on both sides and divided them. So $\frac{20x^2}{-5x^2}$ and because the $x$'s cancel out so I can't use $\infty$ in place of $x$, I'm left with $-4$.
If anyone can verify that I got the right answer, I would appreciate it.
 A: We use l'Hospitals rule for proof:
$$\begin{align*}\lim_{x\to\infty} \frac{20x^2 + 6x + 4}{-5x^2 + 7x - 6} & = \left [ \frac{\infty}{-\infty} \right ] \\
= \lim_{x\to\infty} \frac{40x + 6}{-10x + 7} & = \left [ \frac{\infty}{-\infty} \right ] \\
= \lim_{x\to\infty} \frac{40}{-10} & = -4
\end{align*}$$
The indeterminate Forms are necessary for l'Hospital to hold.
A: You're absolutely on the right track. Note that for non-zero $x,$ we can rewrite $$\frac{20x^2+6x+4}{7x-6-5x^2}=\cfrac{\frac{20x^2+6x+4}{x^2}}{\frac{7x-6-5x^2}{x^2}}=\cfrac{20+\frac6x+\frac4{x^2}}{\frac7x-\frac5{x^2}-5}.$$ The limit of the numerator on the far right as $x\to\infty$ is $20$ and the corresponding limit of the far-right denominator is $-5.$ Hence, since $x$ is eventually non-zero as we let $x$ grow without bound, we have $$\lim_{x\to\infty}\frac{20x^2+6x+4}{7x-6-5x^2}=\frac{20}{-5}=-4,$$ as you said.
A: The key is polynomial division.$$\begin{align*}\lim_{x\to\infty} \frac{20x^2+6x+4}{7x-6-5x^2}&=\lim_{x\to\infty}-\frac{20x^2+6x+4}{5x^2-7x+6}\\&=\lim_{x\to\infty}-4\frac{5x^2+\frac32x+1}{5x^2-7x+6}\\&=\lim_{x\to\infty}\left(-4+O\left(\frac1x\right)\right)\\&=-4\end{align*}$$
