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COMEDK 2025 Evening Solved Paper

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Question : 150 of 180

Marks: +1, -0

Evaluate:

\lim_{x \rightarrow 0} \frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}

1
0
$\frac{2}{3}$
$\frac{1}{3}$

Solution:

Concept:

Use the binomial expansion for fractional powers to simplify the expression when

x

is small.

Explanation:

For small

x

, expand

(1+x)^{1/3}

and

(1-x)^{1/3}

using the binomial series:

(1+x)^{1/3} = 1 + \frac{x}{3} + O(x^2)

(1-x)^{1/3} = 1 - \frac{x}{3} + O(x^2)

Subtract the two expansions:

\sqrt[3]{1+x} - \sqrt[3]{1-x} = \left(1 + \frac{x}{3} + O(x^2)\right) - \left(1 - \frac{x}{3} + O(x^2)\right) = \frac{2x}{3} + O(x^2)

Divide by

x

:

\frac{\sqrt[3]{1+x} - \sqrt[3]{1-x}}{x} = \frac{2}{3} + O(x)

As

x \to 0

, the term

O(x)

approaches

0

, so the limit is

\frac{2}{3}

.

Answer:

Option C:

\frac{2}{3}

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