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Question Number 142922 by mathlove last updated on 07/Jun/21

$$\underset{x\to 1}{\lim }\frac{\sin (x+1)}{2x-\sqrt{x^2+3}}=?$$

Answered by Olaf_Thorendsen last updated on 07/Jun/21

$$\frac{\sin (x+1)}{2x-\sqrt{x^2+3}}=\frac{\sin (x+1).(2x+\sqrt{x^2+3})}{4x^2-(x^2+3)}$$$$=\frac{\sin (x+1).(2x+\sqrt{x^2+3})}{3(x+1)(x-1)}$$$$\underset{1}{\sim }\frac{2\sin \left(2\right)}{3(x-1)}$$$$\underset{x\to 1^{-}}{\lim }\frac{\sin (x+1)}{2x-\sqrt{x^2+3}}=-\mathrm{\infty }$$$$\underset{x\to 1^{+}}{\lim }\frac{\sin (x+1)}{2x-\sqrt{x^2+3}}=+\mathrm{\infty }$$

Answered by mathmax by abdo last updated on 07/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sin (\mathrm{x}+1)}{2\mathrm{x}-\sqrt{{\mathrm{x}}^2+3}}\mathrm{changement}\mathrm{t}=\mathrm{x}-1\mathrm{give}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{t}+1)=\frac{\sin (\mathrm{t}+2)}{2(\mathrm{t}+1)-\sqrt{{(\mathrm{t}+1)}^2+3}}(\mathrm{t}\to 0)$$$$=\frac{\sin (\mathrm{t}+2)}{2\mathrm{t}+2-\sqrt{{\mathrm{t}}^2+2\mathrm{t}+4}}=\frac{\sin (\mathrm{t}+2)}{2\mathrm{t}+2-2\sqrt{1+\frac{{\mathrm{t}}^2+2\mathrm{t}}{4}}}\sim \frac{\sin (\mathrm{t}+2)}{2\mathrm{t}+2-2(1+\frac{1}{8}({\mathrm{t}}^2+2\mathrm{t}))}$$$$=\frac{\sin (\mathrm{t}+2)}{2\mathrm{t}+2-2-\frac{1}{4}{\mathrm{t}}^2-\frac{1}{2}\mathrm{t}}=\frac{\sin (\mathrm{t}+2)}{\frac{3}{2}\mathrm{t}-\frac{1}{4}{\mathrm{t}}^2}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{f}(\mathrm{t}+1)=\mathrm{\infty }={\lim }_{\mathrm{x}\to 1}\mathrm{f}\left(\mathrm{x}\right)$$

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