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AP ICET 2024 6th May 2024 Shift 1

Question : 164 of 200

Marks: +1, -0

\lim\limits_{x \to 0} \frac{\sqrt[3]{8+x}-2}{x}=

Solution:

We are given:

L=\lim\limits_{x \to 0} \frac{\sqrt[3]{8+x}-2}{x}

We apply the binomial expansion for small

x

\sqrt[3]{8+x} \approx 2+\frac{x}{12} \text{ for small } x

Thus, the given expression becomes:

\lim\limits_{x \to 0} \frac{ \left(2+\frac{x}{12}\right) -2}{x} = \lim\limits_{x \to 0} \frac{ \frac{x}{12} }{x} = \frac{1}{12} .

Quick Tip

For cube roots, use the binomial expansion for small values of

x

to simplify the limit expression.

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