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Question Number 96977 by 175 last updated on 05/Jun/20

$$\underset{x\to 5}{\lim }\frac{x-5}{2+{\cot }^2\frac{2}{x-5}}$$

Answered by mathmax by abdo last updated on 06/Jun/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-5}{2+\frac{1}{{\tan }^2\left(\frac{2}{\mathrm{x}-5}\right)}}\mathrm{changement}\mathrm{x}-5=\mathrm{t}\mathrm{give}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{t}\right)$$$$=\frac{\mathrm{t}}{2+\frac{1}{{\tan }^2\left(\frac{2}{\mathrm{t}}\right)}}=\frac{{\mathrm{ttan}}^2\left(\frac{2}{\mathrm{t}}\right)}{{2\tan }^2\left(\frac{2}{\mathrm{t}}\right)+1}(\mathrm{x}\to 5\Rightarrow \mathrm{t}\to 0)$$$$\mathrm{g}\left(\mathrm{t}\right)=\frac{{\mathrm{tsin}}^2\left(\frac{2}{\mathrm{t}}\right)}{{2\sin }^2\left(\frac{2}{\mathrm{t}}\right)+{\cos }^2\left(\frac{2}{\mathrm{t}}\right)}=\frac{{\mathrm{tsin}}^2\left(\frac{2}{\mathrm{t}}\right)}{1+{\sin }^2\left(\frac{2}{\mathrm{t}}\right)}$$$$0⩽{\sin }^2\left(\frac{2}{\mathrm{t}}\right)⩽1\Rightarrow 1⩽1+{\sin }^2\left(\frac{2}{\mathrm{t}}\right)⩽2\Rightarrow \frac{1}{2}⩽\frac{1}{1+{\sin }^2\left(\frac{2}{\mathrm{t}}\right)}⩽1\Rightarrow$$$$\frac{\mid \mathrm{t}\mid }{2}⩽\mid \mathrm{g}\left(\mathrm{t}\right)\mid ⩽\mid \mathrm{t}\mid \to 0(\mathrm{t}\to 0)\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{g}\left(\mathrm{t}\right)=0={\lim }_{\mathrm{x}\to 5}\mathrm{f}\left(\mathrm{x}\right).$$

Commented by 175 last updated on 06/Jun/20

you are a great mathematical thanks alot

Commented by abdomathmax last updated on 06/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

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