\lim_{y\to 0_+}\left (\frac{1}{\ln(y+1)}+2\right )y=\lim_{y\to 0_+}\frac{y+2y\ln(y+1)}{\ln(y+1)}=\lim_{y\to 0_+}\frac{1+2\ln(y+1)+\frac{2y}{y+1}}{\frac{1}{y+1}}=1