Answer:
(B) 3
Explanation:
Qualitative answer. As x grows larger and larger, the contribute of lower degree terms (and constants) becomes irrelevant, and you rewrite everything picking the highest term for both numerator and denominator
![\lim\limits_(x \to \infty) (√(9x^4))/(x^2) $\lim\limits_(x \to \infty) (3x^2)/(x^2)=3](https://img.qammunity.org/2023/formulas/mathematics/college/4r1cptszgt8wm4sked8o5j2bj5moxvyg8r.png)
Analitically.
Collect
in the numerator under the square root,
in the denominator.
![\lim\limits_(x \to \infty) \frac{\sqrt{9x^4(1+(1)/(9x^4))}}{x^2(1-(3)/(x)+(5)/(x^2))} = \lim\limits_(x \to \infty)\frac{3x^2\sqrt{(1+(1)/(9x^4))}}{{x^2(1-(3)/(x)+(5)/(x^2))}} = 3](https://img.qammunity.org/2023/formulas/mathematics/college/rhycnrm1h60ysmeyslj643tb2um1m4gkyf.png)
At this point the way you deal with it is the usual. Anything inside the bracket goes to 0 besides the lead coefficients (1 in both cases) and you're left with 3.