Oops! two limits

note that: $\frac{e^x}{y^2}(\ln((x+y)^y)e^y+\ln(\frac{1}{x^y})) = \frac{e^x}{y^2}(y\ln((x+y))e^y-y\ln(x)) = \frac{e^x\color{#D61F06}y}{y^{\color{#D61F06}2}}(\ln(x+y)e^y-\ln(x))$ $=\frac{e^x(\ln(x+y)e^y-\ln(x))}{y}=\frac{\ln(x+y)e^xe^y-\ln(x)e^x}{y}$ $=\frac{\ln(x+y)e^{x+y}-\ln(x)e^x}{y}$ take the limit: $\lim_{y\to0}\frac{\ln(x+y)e^{x+y}-\ln(x)e^x}{y}$ note that using the definition of the derivative, this limit is the same as $\frac{d}{dx}(\ln(x)e^x)=\ln(x)e^x+\frac{e^x}{x}$ taking the other limit gives: $\lim_{x\to1}(\ln(x)e^x+\frac{e^x}{x})$ $=\ln(1)e^1+\frac{e^1}{1}$ $=0+e=\boxed{e}.$