<s>[INST] Evaluating $\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}$ without L'Hopital I have the following limit question, where different indices of roots appear in the numerator and the denominator
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}.$$
As we not allowed to use L'Hopital, I want to learn how we can proceed algebraically.
 [/INST] A possible purely algebraic way can use 
$$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$$
Using this formula note that


*

*$\sqrt[4]{x-2} - 1 = \frac{\left(\sqrt[4]{x-2}\right)^4 - 1^4 }{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1} =  \frac{x-3 }{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1}$

*$\sqrt[3]{x+5} - 2 = \frac{\left(\sqrt[3]{x+5}\right)^3- 2^3 }{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2} = \frac{x-3 }{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2}$
So, you get
$$\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1} = \frac{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1}{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2} \stackrel{x \to 3}{\longrightarrow}\frac{4}{12} = \frac{1}{3}$$
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