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