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Question Number 145806 by bramlexs22 last updated on 08/Jul/21

$$\underset{x\to \mathrm{\infty }}{\lim }\mathrm{x}({\tan }^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}+4}\right)-\frac{\pi }{4})=?$$

Answered by gsk2684 last updated on 08/Jul/21

$$\underset{x\to \mathrm{\infty }}{\lim }x\left({\tan }^{-1}\left(\frac{\frac{x+1}{x+4}-1}{1+\frac{x+1}{x+4}}\right)\right)$$$$\underset{x\to \mathrm{\infty }}{\lim }x\left({\tan }^{-1}\left(\frac{-3}{2x+5}\right)\right)$$$$\underset{x\to \mathrm{\infty }}{\lim }x\left(\frac{-3}{2x+5}\right)=-\frac{3}{2}$$

Answered by mathmax by abdo last updated on 09/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}(\arctan \left(\frac{\mathrm{x}+1}{\mathrm{x}+4}\right)-\frac{\pi }{4})\Rightarrow \mathrm{f}\left(\mathrm{x}\right){=}_{\mathrm{x}=\frac{1}{\mathrm{t}}}\frac{1}{\mathrm{t}}(\arctan \left(\frac{1+\frac{1}{\mathrm{t}}}{4+\frac{1}{\mathrm{t}}}\right)-\frac{\pi }{4})$$$$=\frac{1}{\mathrm{t}}(\arctan \left(\frac{\mathrm{t}+1}{4\mathrm{t}+1}\right)-\frac{\pi }{4})=\mathrm{z}(\mathrm{twe}\mathrm{have}$$$$\arctan \left(\frac{\mathrm{t}+1}{4\mathrm{t}+1}\right)=\arctan (\frac{1}{4}\left(\frac{4\mathrm{t}+1+3}{4\mathrm{t}+1}\right)=\arctan (\frac{1}{4}+\frac{3}{4(4\mathrm{t}+1)})$$$$\sim \arctan (\frac{1}{4}+\frac{3}{4}(1-4\mathrm{t}))=\arctan (1-3\mathrm{t})$$$$\mathrm{we}\mathrm{have}\tan (\arctan (1-3\mathrm{t})-\frac{\pi }{4})=\frac{1-3\mathrm{t}-1}{1+1-3\mathrm{t}}=\frac{-3\mathrm{t}}{2-3\mathrm{t}}$$$$=>\mathrm{z}\left(\mathrm{t}\right)\sim \frac{1}{\mathrm{t}}\arctan \left(\frac{-3\mathrm{t}}{(2-3\mathrm{t})}\right)\sim =\frac{1}{\mathrm{t}}\frac{-3\mathrm{t}}{2-3\mathrm{t}}\to -\frac{3}{2}(\mathrm{t}\to 0)\Rightarrow$$$${\lim }_{\mathrm{x}\to \mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=-\frac{3}{2}$$

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