How do I do this?
lim (square root(8x^3 + 5x + 10)) / (x^2)
x->infinity
Any help?
$\displaystyle \lim_{x \to \infty} \frac{\sqrt{8x^3 + 5x + 10}}{x^2} \cdot \frac{\displaystyle {\color{red}\frac{1}{x^2}}}{\ \displaystyle{\color{red}\frac{1}{x^2}} \ } \ = \ \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{x^2}}{\displaystyle \frac{\ x^2 \ }{x^2}}$
$\displaystyle = \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{\sqrt{x^4}}}{1} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3 + 5x + 10}{x^4}} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3}{x^4} + \frac{5x}{x^4} + \frac{10}{x^4}} \ = \ \hdots$
Not exactly. $\displaystyle \sqrt{x^3} = x^{\frac{3}{2}} \ < \ x^2$