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Question Number 40092 by maxmathsup by imad last updated on 15/Jul/18

$$calculate{lim}_{x\to +\mathrm{\infty }}\frac{1}{x}tan\left(\frac{\pi x}{2x+3}\right)$$

Commented by math khazana by abdo last updated on 26/Jul/18

$$wehave\frac{\pi x}{2x+3}=\frac{\pi x}{2x(1+\frac{3}{2x})}=\frac{\pi }{2}.\frac{1}{1+\frac{3}{2x}}$$$$\sim \frac{\pi }{2}(1-\frac{3}{2x}+0\left(\frac{1}{x}\right))(x\to +\mathrm{\infty })\Rightarrow$$$$tan\left(\frac{\pi x}{2x+3}\right)\sim tan(\frac{\pi }{2}-\frac{3\pi }{4x})\sim \frac{1}{tan\left(\frac{3\pi }{4x}\right)}\Rightarrow$$$$\frac{1}{x}tan\left(\frac{\pi x}{2x+3}\right)\sim \frac{1}{xtan\left(\frac{3\pi }{4x}\right)}changement$$$$\frac{3\pi }{4x}=tgive4tx=3\pi \Rightarrow {lim}_{x\to +\mathrm{\infty }}\frac{1}{xtan\left(\frac{3\pi }{4x}\right)}$$$$={lim}_{t\to 0}\frac{1}{\frac{3\pi }{4t}tan\left(t\right)}=\frac{4}{3\pi }{lim}_{t\to 0}\frac{1}{\frac{tant}{t}}=\frac{4}{3\pi }$$$$because{lim}_{t\to 0}\frac{tant}{t}=1$$

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