<s>[INST] Where is the mistake in this limit calculation? I got this limit:
$$\lim_{x\to1}\frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}*\frac{1}{\sqrt[4]{x}-1}*\frac{\sqrt[4]{x}+1}{\sqrt[4]{x}+1} \implies \lim_{x\to1}\frac{\sqrt[4]{x}+1}{\sqrt[3]{x²}+\sqrt[3]{x}+1} \xrightarrow[x\to 1]{}\frac{2}{3}.$$
But in the book the answer is $$ \frac{4}{3}.$$
I cannot find the mistake in my calculation.
 [/INST] If fractional exponents are difficult to work with, you can also try a variable substitution:  let $x = u^{12}$, so that $\sqrt[3]{x} = u^4$ and $\sqrt[4]{x} = u^3$.  Then we have $$\frac{\sqrt[3]{x} - 1}{\sqrt[4]{x} - 1} = \frac{u^4 - 1}{u^3 - 1} = \frac{(u-1)(u+1)(u^2+1)}{(u-1)(u^2+u+1)} = \frac{(u+1)(u^2+1)}{u^2+u+1}.$$  And since $x \to 1$ implies $u \to 1$, the limit is simply $$\frac{(1+1)(1^2+1)}{1^2+1+1} = \frac{4}{3},$$ as claimed.
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