Q72 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
Evaluate: $\lim _{x \rightarrow 0} \dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}$
- Correct answerA$\dfrac{2}{3}$
- B$\dfrac{1}{3}$
- C$0$
- D$1$
Solution
Let $f(x) = \dfrac{(1+x)^{1/3} - (1-x)^{1/3}}{x}$.
Using the binomial expansion $(1+u)^n = 1 + nu + \dfrac{n(n-1)}{2!}u^2 + \dots$ for small $u$, we have:
$(1+x)^{1/3} = 1 + \dfrac{1}{3}x + \dfrac{\dfrac{1}{3}(\dfrac{1}{3}-1)}{2}x^2 + \dots = 1 + \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dots$
$(1-x)^{1/3} = 1 - \dfrac{1}{3}x + \dfrac{\dfrac{1}{3}(\dfrac{1}{3}-1)}{2}(-x)^2 + \dots = 1 - \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dots$
Substituting these into the expression:
$\lim _{x \rightarrow 0} \dfrac{(1 + \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dots) - (1 - \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dots)}{x}$
$\lim _{x \rightarrow 0} \dfrac{\dfrac{2}{3}x + O(x^3)}{x}$
$\lim _{x \rightarrow 0} (\dfrac{2}{3} + O(x^2)) = \dfrac{2}{3}$
Answer: $\dfrac{2}{3}$
Using the binomial expansion $(1+u)^n = 1 + nu + \dfrac{n(n-1)}{2!}u^2 + \dots$ for small $u$, we have:
$(1+x)^{1/3} = 1 + \dfrac{1}{3}x + \dfrac{\dfrac{1}{3}(\dfrac{1}{3}-1)}{2}x^2 + \dots = 1 + \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dots$
$(1-x)^{1/3} = 1 - \dfrac{1}{3}x + \dfrac{\dfrac{1}{3}(\dfrac{1}{3}-1)}{2}(-x)^2 + \dots = 1 - \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dots$
Substituting these into the expression:
$\lim _{x \rightarrow 0} \dfrac{(1 + \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dots) - (1 - \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dots)}{x}$
$\lim _{x \rightarrow 0} \dfrac{\dfrac{2}{3}x + O(x^3)}{x}$
$\lim _{x \rightarrow 0} (\dfrac{2}{3} + O(x^2)) = \dfrac{2}{3}$
Answer: $\dfrac{2}{3}$
Hence, the correct answer is option: 'A'.