<s>[INST] Solve $\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}$ I tried first without L'Hôpital's rule:
$$\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}} = 
\frac{\sqrt{x}}{\sqrt[3]{x}} \cdot \frac{1+\frac{x}{\sqrt x}}{1+ \sqrt[3] x} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\frac{\sqrt x+x}{\sqrt x}}{\frac{\sqrt[3]{x}+x}{\sqrt[3]{x}}} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\sqrt[3]{x}(\sqrt x +x)}{\sqrt x (\sqrt[3]{x}+x)} = \frac{\sqrt x \sqrt[3]{x}(\sqrt x +x)}{\sqrt[3] x \sqrt x (\sqrt[3]x +x )} = \frac{\sqrt x +x }{\sqrt[3] x +x }$$
That didn't work, so then I tried L'Hôpital's:
$$\frac{\frac{1}{2}(x)^{-\frac{1}{2}}+1}{1+\frac{1}{3}(x)^{-\frac{1}{3}}} = \frac{\frac{1}{2\sqrt x }+1}{1+\frac{1}{{3\sqrt[3]{x}}}} = \frac{\frac{1+2\sqrt x}{2\sqrt x}}{\frac{1+3\sqrt[3]{x}}{3\sqrt[3]{x}}} = \frac{3\sqrt[3]{x}(1+2\sqrt x)}{2\sqrt x (1+3\sqrt[3]{x})} = \frac{3\sqrt[3]{x}+3\sqrt[3]{x}2\sqrt{x}}{2\sqrt{x} + 2\sqrt{x}3\sqrt[3]{x}} = ??? $$
How do I solve this? If you can solve this with and without L'Hôpital's, please do so.
 [/INST] In such cases, it is easier to perform a suitable transformation to obtain integer-valued powers:  this suggests letting $x = u^6$, hence we have the equivalent limit $$\lim_{x \to \infty} \frac{x^{1/2} + x}{x+x^{1/3}} = \lim_{u \to \infty} \frac{u^3 + u^6}{u^6 + u^2} = \lim_{u \to \infty} \frac{u^{-3} + 1}{1 + u^{-4}}$$ and the rest is trivial.
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