$\lim\limits_{x\to\infty} \dfrac{\sqrt{9x^4-2}}{4x^2-2x+5}$

Summary Submit Skip Question

When determining limits at infinity, only the highest power terms in the numerator and denominator are relevant.

$$\lim\limits_{x\to\infty} \dfrac{\sqrt{9x^4-2}}{4x^2-2x+5} \approx \lim\limits_{x\to\infty} \frac{\sqrt{9x^4}}{4x^2}$$ $$= \lim\limits_{x\to\infty}\frac{3x^2}{4x^2}$$ $$= \lim\limits_{x\to\infty}\frac{3}{4}$$ $$= \boxed{\frac{3}{4}}$$

We can divide the numerator and denominator by $x^2$.

$$\lim\limits_{x\to\infty} \dfrac{\sqrt{9-\frac{2}{x^4}}}{4-\frac{2}{x}+\frac{5}{x^2}}$$

Recall that as $x$ approaches large values, fractions in the form of $\dfrac{c}{x^r}$ approach 0.

$$= \frac{\sqrt{9-0}}{4-0+0}$$ $$= \boxed{\frac{3}{4}}$$