Let a function U(T) is given by, $U(T)=\frac{1×e^{-\frac{\epsilon}{k_BT}}+2×e^{-\frac{2\epsilon}{k_BT}}+3×e^{-\frac{3\epsilon}{k_BT}}}{1×e^{-\frac{\epsilon}{k_BT}}+1×e^{-\frac{2\epsilon}{k_BT}}+1×e^{-\frac{3\epsilon}{k_BT}}}$; where $\epsilon$ and $k_B$ are constants and T is absolute Temperature (in Kelvin Scale).
When I try to plot T vs U(T) we get 
as we can see from the plot, U(T=0K)=1. But if I try to get it analitically i.e. $$ \lim _{T \rightarrow 0} U(T)=\lim _{T \rightarrow 0}\frac{1×e^{-\frac{\epsilon}{k_BT}}+2×e^{-\frac{2\epsilon}{k_BT}}+3×e^{-\frac{3\epsilon}{k_BT}}}{1×e^{-\frac{\epsilon}{k_BT}}+1×e^{-\frac{2\epsilon}{k_BT}}+1×e^{-\frac{3\epsilon}{k_BT}}}=\frac{0}{0} form $$ -this can't even be solved by using L'Hôpital's Rule because the $\frac{0}{0}$ form comes again and again.
the How can I proove U(T=0K)=1 ?
Assuming $\epsilon>0$ and $k_B>0$, with $y=\exp(-\epsilon/(k_BT))$ we have $$ \lim_{y\to0}\frac{y+2y^2+3y^3}{y+y^2+y^3}=\lim_{y\to0}\frac{1+2y+3y^2}{1+y+y^2}=1 $$