Salut,
Cazul de nedeterminare este ∞⁰ (infinit la puterea 0).
Soluția este e, vezi mai jos:
[tex]f(e^x)=\dfrac{2(e^x)^3}{(e^x)^2+1}=\dfrac{2e^{3x}}{e^{2x}+1}=\dfrac{2e^{3x}}{e^{2x}\left(1+\dfrac{1}{e^{2x}}\right)}=\dfrac{2e^{3x-2x}}{1+\dfrac{1}{e^{2x}}}=\dfrac{2e^x}{1+\dfrac{1}{e^{2x}}}.\\\\\\\left[f(e^x)\right]^{\frac{1}x}=\left[\dfrac{2e^x}{1+\dfrac{1}{e^{2x}}}\right]^{\frac{1}x}=e\cdot\left(\dfrac{2}{1+\dfrac{1}{e^{2x}}}\right)^{\frac{1}x}\to\ e\cdot\left(\dfrac{2}{1+0}\right)^0=e\cdot 1=e.[/tex]
Green eyes.
