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Question Number 204396 by Thierrybadouana last updated on 15/Feb/24

Answered by Faetmaaa last updated on 27/Feb/24

$$\ln (1+y)\underset{y\to 0}{\sim }y$$$$\sin \left(y\right)\underset{y\to 0}{\sim }y$$$$\underset{\begin{array}{c}x\to 0\\ x\ne 0\end{array}}{\lim }\frac{\ln (1+x^2)}{{\sin }^2\left(x\right)}=\underset{\begin{array}{c}x\to 0\\ x\ne 0\end{array}}{\lim }\frac{x^2}{x^2}=\underset{\begin{array}{c}x\to 0\\ x\ne 0\end{array}}{\lim }1=1$$

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