<s>[INST] Evaluate $\lim_{x \to 4}\frac{(2x)^{1/3} - 2}{\sqrt{x} - 2}$ without L'Hospital rule I know how to solve this limit with the L'Hospital rule:
$$\lim_{x \rightarrow 4}  \frac{(2x)^{1/3} - 2}{\sqrt{x} -2}$$
The answer is $\frac{2}{3}$. I am trying to solve it without the L'Hospital rule.
 [/INST] $$\frac{(2x)^{1/3}-2}{\sqrt x-2}=\sqrt[3]2\frac{\sqrt[3]x-\sqrt[3]{2^2}}{\sqrt x-2}=\sqrt[3]2\frac{\overbrace{x-4}^{=(\sqrt x-2)(\sqrt x+2)}}{x^{2/3}+\sqrt[3]{4x}+4^{2/3}}\frac1{\sqrt x-2}=$$
$$=\frac{\sqrt[3]2(\sqrt x+2)}{x^{2/3}+\sqrt[3]{4x}+4^{2/3}}\xrightarrow[x\to 4]{}\frac{\sqrt[3]2\cdot4}{3\cdot2\sqrt[3]2}=\frac23$$
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