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Question Number 204395 by Thierrybadouana last updated on 15/Feb/24

Answered by Lindemann last updated on 20/Feb/24

$$=\underset{x\to 0}{\lim }\left(\frac{ln(1+x^2)}{x^2}\times \frac{x^2}{{sin}^2\left(x\right)}\right)=1$$

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