Q27 / 90 · NTA Abhyas (JEE Main) 2020 · Mathematics · Limits

The value of $\underset{x \to 1}{l i m} \frac{\sqrt[5]{x^{2}} - 2 \sqrt[5]{x} + 1}{{(x - 1)}^{2}}$ is equal to

Correct answer: 0.04

Solution

\underset{x \to 1}{l i m} \frac{\sqrt[5]{x^{2}} - 2 \sqrt[5]{x} + 1}{{(x - 1)}^{2}} = \underset{x \to 1}{l i m} \frac{y^{2} - 2 y + 1}{{(y^{5} - 1)}^{2}}

[P u t t i n g \sqrt[5]{x} = y, as x \to 1, y \to 1]

\underset{y \to 1}{l i m} \frac{{(y - 1)}^{2}}{{(y - 1)}^{2} {(y^{4} + y^{3} + y^{2} + y + 1)}^{2}}

= \frac{1}{25}

Hence, the correct answer is 0.04.

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