I need to solve
$$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{x-1} $$
But without using L’Hopital’s Rule, could someone help me?
2026-03-28 10:56:23.1774695383
Limits with cubic root without L’Hopital’s Rule: $\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{x-1} $
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Factorize $x-1$ as $$\left(\sqrt[3]{x}-1\right)\left(x^{2/3}+x^{1/3}+1\right)$$