Evaluate $\lim_{x \to 0}{\frac{\sqrt[3]{1 + cx} - 1}{x}}$ without using L-H Rule [duplicate]

limits limits-without-lhopital

Hint: Multiply the numerator and denominator by $\sqrt[3]{1+cx}^2+\sqrt[3]{1+cx}+1$.

This is based on the idea that $(a-b)(a^2+ab+b^2)=a^3-b^3$. We use it with $a=\sqrt[3]{1+cx}$ and $b=1$, transforming the numerator into $cx$.

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