Find the value of : $\lim_{x \rightarrow \infty} \sqrt{(8x^2-3)/(2x^2+x)} $

Question

$$ \lim_{x \rightarrow \infty} \sqrt{\frac{8x^2-3}{2x^2+x}} $$ Do I square the whole function? I don't know how to start.

Answer 1

Remember this fact: $(n > 0 )$

$$ \lim_{x \to \infty } \frac{1}{x^n} = 0 $$

Now, notice

$$ \sqrt { \frac{ 8x^2 - 3}{2x^2 + x } } = \sqrt{ \frac{8 - \frac{3}{x^2}}{2 + \frac{1}{x}}} \to_{x \to \infty} \sqrt{ \frac{8}{2} } = \sqrt{4} = 2$$

Answer 2

It is equal:

$$\sqrt{\lim_{x\to \infty}\frac{8x^2-3}{2x^2+x}}=\sqrt{4}=2$$

Answer 3

Hint 1:

$$\frac{\sqrt{a}}{x} = \sqrt{\frac{a}{x^2}} $$