$$\lim_{x \to \infty} \frac{\sqrt[3]{x+2}}{\sqrt{x+3}} $$
How would you proceed to find this limit, by eyeballing I would guess it foes to zero since the numerator has a smaller power than the denominator, normaly I would use the binomial theorem if I had something like $$\lim_{x \to \infty} \frac{\sqrt[3]{x+2}-1}{\sqrt{x+3}-1} $$ But here I don't know how to find the limit since I can't really use the binomial theorem.
Best Answer
$$\lim_{x \to \infty} \frac{\sqrt[3]{x+2}}{\sqrt{x+3}}=\lim_{x \to \infty} \sqrt[6]{\dfrac{(x+2)^2}{(x+3)^3}}=\sqrt[6]{0}=0$$