Practice: Limits.

The exponent of n in the numerator is greater than the exponent of n in the denominator. Also, the power of n in the denominator is negative. Thus, the limit approaches negative infinity.

Since we have a $-\infty/\infty$ case, we can use L'Hôpital's rule: $\begin{aligned} \lim\limits_{n\to\infty}\dfrac{17n-2}{9-\sqrt{n}}&=\lim\limits_{n\to\infty}\dfrac{17-0}{0-1/\sqrt{4n}}\\ &=\lim\limits_{n\to\infty}-34\sqrt{n}\\ &=-\infty. \end{aligned}$

Exactly as that guy said it $\uparrow$ . First, we can evaluate the numerator and denominator separately, and when it's combined, it becomes $\frac{\infty}{-\infty}$ . We can apply L'Hopital's, and just as above, we attain $-\infty$ . Another way of thinking about it: what's a positive divided by a negative? It's negative, and we're done.