Q33 / 90 · NTA Abhyas (JEE Main) 2020 · Mathematics · Limits

The value of $\underset{x \to 0}{l i m} \frac{l n (10 - 9 \cos 2 x)}{l n^{2} (\sin 3 x + 1)}$ is equal to

Correct answer: 2

Solution

\underset{x \to 0}{l i m} \frac{l n (10 - 9 \cos 2 x)}{l n^{2} (\sin 3 x + 1)} = \underset{x \to 0}{l i m} \frac{l n \{1 + (9 - 9 \cos 2 x)\}}{l n^{2} (1 + \sin 3 x)}

= \underset{x \to 0}{l i m} \frac{9 (1 - \cos 2 x)}{{(\sin 3 x)}^{2}} = \underset{x \to 0}{l i m} \frac{9 (2 x^{2})}{{(3 x)}^{2}} = 2

Hence, the correct answer is 2.

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