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. Divide through by $\displaystyle x^2$ to get
$\displaystyle \left| \frac{9}{x^2} + \frac{5}{x} - 6 \right| \big/ \left( 2 - \frac{\sin x}{x} \right)$.
Now, both the numerator and denominator have a finite limit.

edit: Since $\displaystyle |\sin x|\leq1$, $\displaystyle \left| \sin x \over x \right | \leq \frac 1 x$. Does that help?

3. Yes i got it. using squeeze theorm x-> infinity for sinx/x is 0.

4. You posted this same question under "PreCalculus". Please do not post questions more than once.

5. Originally Posted by HallsofIvy
You posted this same question under "PreCalculus". Please do not post questions more than once.
Yep. sorry but i've already renamed that thread so that the mods could delete it. Only found out i posted in the wrong thread after i submitted my question.