Question

Evaluate $\lim_{x\to 0}\left( \frac{1-\sqrt[3]{x^2+1}}{x^2} \right)$

Answer 1

The limit evaluates to $-\frac{1}{3}$.

Answer 2

The limit is of the indeterminate form $\frac{0}{0}$. We used algebraic manipulation involving the difference of cubes identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ to rationalize the numerator. By setting $a=1$ and $b=(x^2+1)^{1/3}$, we multiplied the numerator and denominator by $a^2+ab+b^2 = 1+(x^2+1)^{1/3}+(x^2+1)^{2/3}$. This transformed the numerator into $a^3-b^3 = 1-(x^2+1) = -x^2$. After canceling $x^2$ from the numerator and denominator (for $x \neq 0$), the expression simplified to $\frac{-1}{1+(x^2+1)^{1/3}+(x^2+1)^{2/3}}$. Evaluating the limit as $x \to 0$ by direct substitution gives $\frac{-1}{1+1+1} = -\frac{1}{3}$.

Alternatively, L'Hopital's Rule could be applied.