<s>[INST] Find the value of : $\lim_{x \rightarrow \infty} \sqrt{(8x^2-3)/(2x^2+x)} $ $$
\lim_{x \rightarrow \infty} \sqrt{\frac{8x^2-3}{2x^2+x}}
$$
Do I square the whole function? I don't know how to start.
 [/INST] Remember this fact: $(n > 0 )$
$$ \lim_{x \to \infty } \frac{1}{x^n} = 0 $$
Now, notice 
$$ \sqrt { \frac{ 8x^2 - 3}{2x^2 + x } } = \sqrt{ \frac{8 - \frac{3}{x^2}}{2 + \frac{1}{x}}} \to_{x \to \infty} \sqrt{ \frac{8}{2} } = \sqrt{4}  = 2$$
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