Question-204396

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 (y) \underset{y \to 0}{\sim} y

\lim_{\begin{matrix} x \to 0 \\ x \neq 0 \end{matrix}} \frac{\ln (1 + x^{2})}{\sin^{2} (x)} = \lim_{\begin{matrix} x \to 0 \\ x \neq 0 \end{matrix}} \frac{x^{2}}{x^{2}} = \lim_{\begin{matrix} x \to 0 \\ x \neq 0 \end{matrix}} 1 = 1

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