What is the Limit?

A property of logarithms: $log(x)+log(y)=log(x\cdot y)$ . That being said: $log(2+\frac{1}{x^2})+log(\frac{1}{4+\frac{1}{x^2}})=log(2+\frac{1}{x^2}\cdot{\frac{1}{4+\frac{1}{x^2}}})$ This becomes: $log(\frac{2+\frac{1}{x^2}}{4+\frac{1}{x^2}})$ . We can find the number by taking the limit as x approaches infinity: $\lim_{x\rightarrow \infty}\hspace{1mm}log(\frac{2+\frac{1}{x^2}}{4+\frac{1}{x^2}})$ As $x\rightarrow \infty$ , $\frac{1}{x^2} \rightarrow 0$ . Therefore, we are left with: $log(\frac{2}{4})=log(0.5)\approx \hspace{0.5mm} -0.301$