<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] You already get many answers, so I shall try something different hoping that you know Taylor series (if this is not the case, just discard my answer).
Built at $x=b$, you have by Taylor $$x^a=b^a+a b^{a-1} (x-b)+\frac{1}{2} (a-1) a b^{a-2} (x-b)^2+O\left((x-b)^3\right)$$ So, the expansion of the numerator is $$\frac{x-4}{6}-\frac{1}{72} (x-4)^2+O\left((x-4)^3\right)$$ and for the denominator $$\frac{x-4}{4}-\frac{1}{64} (x-4)^2+O\left((x-4)^3\right)$$ Performing the long division, you will the get that, close to $x=4$$$\frac{(2x)^{1/3} - 2}{\sqrt{x} -2}\approx \frac{2}{3}-\frac{x-4}{72}$$ which shows both the limit and how it is approached.
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