Question

lim [tg (pi *x/(2x+1) )]^ (1/x) = ??? ; unde "^" = "la puterea"
x->infinit

Answer 1

[tex]\lim_{x\rightarrow\infty}(\tan\frac{\pi x}{2x+1})^{\frac{1}{x}}=\lim_{x\rightarrow\infty}e^\ln(\tan\frac{\pi x}{2x+1})^{\frac{1}{x}}}=\lim_{x\rightarrow\infty}e^{\frac{1}{x}\ln(\tan\frac{\pi x}{2x+1})}=\\ e^\lim_{x\rightarrow\infty}\frac{\ln(\tan\frac{\pi x}{2x+1})}{x}}\ \\ \lim_{x\rightarrow\infty}\frac{\ln(\tan\frac{\pi x}{2x+1})}{x}=\lim_{x\rightarrow\infty}\frac{[\ln(\tan\frac{\pi x}{2x+1})]'}{x'}=\\ [/tex]
[tex]=\lim_{x\rightarrow\infty}[\frac{1}{\tan\frac{\pi x}{2x+1}}\cdot\frac{1}{\cos^2\frac{\pi x}{2x+1}}\frac{\pi(2x+1)-2\pi x}{(2x+1)^2}]=\\ =\lim_{x\rightarrow\infty}\frac{1}{\tan\frac{\pi x}{2x+1}}\frac{1}{\cos^2\frac{\pi x}{2x+1}}\frac{1}{(2x+1)^2}=\\ =\lim_{x\rightarrow\infty}\frac{1}{\sin\frac{\pi x}{2x+1}\cos\frac{\pi x}{2x+1}\cdot(2x+1)^2}=\\ =\lim_{x\rightarrow\infty}\frac{2}{\sin\frac{2\pi x}{2x+1}(2x+1)^2}=\lim_{x\rightarrow\infty}\frac{2}{\sin\frac{2\pi x}{2x+1}(2x+1)^2}=[/tex]
[tex]\lim_{x\rightarrow\infty}\frac{\frac{2}{(2x+1)^2}}{\sin\frac{2\pi x}{2x+1}}=\\
\lim_{x\rightarrow\infty}\frac{\frac{-4}{(2x+1)^3}}{\cos\frac{2\pi x}{2x+1}\cdot\frac{2\pi}{(2x+1)^2}}=0[/tex]
In final limita este e^0=1