Question

determine the following limit in simplest form. if the limit is infinite, state that the limit does not exist (dne).
lim_{x
ightarrowinfty}\frac{sqrt{41x^{5}+64x^{6}-2x}}{9x + 5x^{3}}

Explanation:

Step1: Identify highest - power terms

For the numerator $\sqrt{41x^{5}+64x^{6}-2x}$, as $x\to\infty$, the highest - power term is $\sqrt{64x^{6}} = 8x^{3}$ (since $64x^{6}$ dominates $41x^{5}$ and $- 2x$ as $x\to\infty$). For the denominator $9x + 5x^{3}$, the highest - power term is $5x^{3}$ as $x\to\infty$.

Step2: Find the limit

We find the limit of the ratio of the highest - power terms. $\lim_{x\to\infty}\frac{\sqrt{41x^{5}+64x^{6}-2x}}{9x + 5x^{3}}\approx\lim_{x\to\infty}\frac{8x^{3}}{5x^{3}}$.

Step3: Simplify the limit expression

$\lim_{x\to\infty}\frac{8x^{3}}{5x^{3}}=\frac{8}{5}$ (the $x^{3}$ terms cancel out).

Answer:

$\frac{8}{5}$