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Question Number 158175 by mathocean1 last updated on 31/Oct/21

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{sinx+x}{3+2sinx}=?$$

Commented by cortano last updated on 01/Nov/21

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sin x+x}{3+2\sin x}=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{\sin x}{x}+1}{\frac{3}{x}+\frac{2\sin x}{x}}$$$$=\frac{0+1}{0+0}=\mathrm{\infty }$$

Answered by puissant last updated on 31/Oct/21

$$u=\frac{1}{x};u\to 0quandx\to +\mathrm{\infty }$$$$\Rightarrow \mathcal{L}=\underset{x\to +\mathrm{\infty }}{\lim }\frac{sinx+x}{3+2sinx}=\underset{u\to 0}{\lim }\frac{sin\left(\frac{1}{u}\right)+\frac{1}{u}}{3+2sin\left(\frac{1}{u}\right)}$$$$=\underset{u\to 0}{\lim }\frac{\frac{1}{u}+\frac{1}{u}}{3+\frac{2}{u}}(enutilisantledevellopement$$$$limit\acute{e}dusinusa{l}^{\prime }ordre1..$$$$\Rightarrow \mathcal{L}=\underset{u\to 0}{\lim }\frac{\frac{2}{u}}{\frac{3u+2}{u}}=\underset{u\to 0}{\lim }\frac{2}{3u+2}=1$$

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