Solving Math Limits Questions with Square Roots I seek your help on this question.
$$\lim_{x\to\infty}\left(\sqrt{9x^6+7x^3+5x}-\sqrt{9x^6+6x^2+3x}\right)$$
This is what I have done so far.

I tried to divide both numerator and denominator with the largest power but I am not too sure if I am doing it right. Please help! Thank you!

We know that, by asymptotic relations:
$$(1+f(x))^{\alpha} \;\; \sim \;\; \alpha \cdot f(x) \;\; \text{when}\;\; f(x) \to 0$$
One can write:
$$\lim_{x\to\infty}\left(\sqrt{9x^6+7x^3+5x}-\sqrt{9x^6+6x^2+3x}\right)=\lim_{x\to\infty}\sqrt{9x^6+6x^2+3x}\cdot\left(\sqrt{\frac{9x^6+7x^3+5x}{9x^6+6x^2+3x}}-1\right)=\lim_{x\to\infty}\sqrt{9x^6+6x^2+3x}\cdot\left(\sqrt{\frac{9x^6+6x^2+3x+(7x^3-6x^2)}{9x^6+6x^2+3x}}-1\right)=\lim_{x\to\infty}\sqrt{9x^6+6x^2+3x}\cdot\left(\sqrt{1+\frac{7x^3-6x^2}{9x^6+6x^2+3x}}-1\right)\;\; \sim\;\;\lim_{x\to\infty}3x^3\cdot\left(\sqrt{1+\frac{7x^3}{9x^6}}-1\right)\;\; \lim_{x\to\infty}3x^3\cdot\frac{1}{2}\cdot\frac{7}{9x^3}=\frac{7}{6}$$
So, the limit is $\frac{7}{6}$.
