lim x-> infinity [ |9+5x-6x^2| ] / [ 2x^2 - xsinx ]

i get 0 as answer. Firstly I used Hopital rule, and then i factor out x^2 before getting 0 as answer. But i think im wrong, can any one help?

I did not just factor out x^2 initially, because if i do that, for the xsinx term in the denominator, it becomes (sinx)/x. And if you take x-> infinity to (sinx)/x, doesnt it give u infinity/infinity and not 0?

2. Originally Posted by toffeefan
lim x-> infinity [ |9+5x-6x^2| ] / [ 2x^2 - xsinx ]

i get 0 as answer. Firstly I used Hopital rule, and then i factor out x^2 before getting 0 as answer. But i think im wrong, can any one help?

I did not just factor out x^2 initially, because if i do that, for the xsinx term in the denominator, it becomes (sinx)/x. And if you take x-> infinity to (sinx)/x, doesnt it give u infinity/infinity and not 0?
No, it doesn't. sin(x) does NOT go to infinity- it stays between 1 and -1. (sin x)/x definitely goes to 0 as x goes to infinity. My recommendation, go ahead and divide both numerator and denominator by $x^2$ as you say. That gives
$\frac{|\frac{9}{x^2}+ \frac{5}{x}- 6|}{2- \frac{sin x}{x}}$ and now everything that still involves an "x" will go to 0.