Răspuns:
⇆[tex]\lim_{x \to \infty} (\frac{x+1}{2x+3}) ^x^{2}[/tex]
Adui si scazi 1 la la baza
lim(1-1+[tex](\frac{x+1}{2x+3} )^x^{2}[/tex]=
lim(1+[tex]\frac{2x+1-2x-3}{2x+3} )^x^2[/tex]=
lim(1-[tex](\frac{2}{2x+3} )^x^2[/tex] ridici baza concomitent la puterea [tex]\frac{-2}{2x+3} *\frac{2x+3}{-2} =1[/tex]
obtii[(1-[tex][(\frac{2}{2x+3} )^\frac{2x+3}{-2} ]^\frac{-2}{2x+3} ^x^2[/tex]
in paranteza dreapta , la limita este e⁻¹
e[tex]-lim\frac{x^2}{2x+3}[/tex]=e^-∞=0
Explicație pas cu pas:
