<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] Make the change: $x-2=t^4$. Then:
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}=\lim_{t\to1}\frac{\sqrt[3]{t^4+7}-2}{t-1}=\\
\lim_{t\to1}\frac{\sqrt[3]{t^4+7}-2}{t-1}\cdot \frac{\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2}{\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2}=\\
\lim_{t\to1} \frac{(t^4+7)-2^3}{(t-1)(\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2)}=\\
\lim_{t\to1} \frac{\require{cancel} \cancel{(t-1)}(t+1)(t^2+1)}{\cancel{(t-1)}(\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2)}=\frac13.$$
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