Răspuns:
Este cazul [tex]1^{\infty}[/tex].
[tex]\displaystyle\lim_{x\to 0}\left(\frac{1+2^x}{2}\right)^{\displaystyle\frac{1}{x}}=\lim_{x\to 0}\left(1+\frac{1+2^x}{2}-1\right)^{\displaystyle\frac{1}{x}}=\\=\lim_{x\to 0}\left[\left(1+\frac{2^x-1}{2}\right)^{\displaystyle\frac{2}{2^x-1}}\right]^{\displaystyle\frac{2^x-1}{2}\cdot\frac{1}{x}}=\\=e^{\displaystyle\frac{1}{2}\lim_{x\to 0}\frac{2^x-1}{x}}=e^{\displaystyle\frac{1}{2}\ln 2}=e^{\ln\sqrt{2}}=\sqrt{2}[/tex]
Explicație pas cu pas:
