Strategy: First simplify expressions, then take limits.
Problem: Find the limits as R→∞R\rightarrow\infty of
Solution. (i) log(2R+1)-log(R+2)=log(2R+1R+2)=log(2+1R1+2R).\log(2R+1)-\log(R+2)=\,{\log\left(\frac{2R+1}{R+2}\right)}{=\log\left(\frac{2+% \frac{1}{R}}{1+\frac{2}{R}}\right).}
Since 2+1R1+2R→21=2\frac{2+\frac{1}{R}}{1+\frac{2}{R}}\rightarrow\frac{2}{1}=2 as R→∞R\rightarrow\infty, we get: limR→∞f(R)=log2\lim_{R\rightarrow\infty}f(R)=\log 2.
(ii) 2log(R+1)-log(3R2+1)=log((R+1)23R2+1)=log((1+1R)23+1R2).2\log(R+1)-\log(3R^{2}+1)=\,{\log\left(\frac{(R+1)^{2}}{3R^{2}+1}\right)=}\,{% \log\left(\frac{(1+\frac{1}{R})^{2}}{3+\frac{1}{R^{2}}}\right).}
As (1+1R)23+1R2→13\frac{(1+\frac{1}{R})^{2}}{3+\frac{1}{R^{2}}}\rightarrow\frac{1}{3} as R→∞R\rightarrow\infty, we obtain: limR→∞g(R)=-log3\lim_{R\rightarrow\infty}g(R)=-\log 3.