## Calculus (3rd Edition)

Take $f(x)=\ln x$ and $g(x)=\ln \frac{1}{x}$. Now, the limits $$\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\ln x, \quad \lim\limits_{x \to 0}g(x)= \lim\limits_{x \to 0}\ln \frac{1}{x}$$ do not exist. Compare to the limit of the sum: $$\lim\limits_{x \to 0}f(x)+ \lim\limits_{x \to 0}g(x)= \lim\limits_{x \to 0}\ln x+\lim\limits_{x \to 0}\ln \frac{1}{x} =\lim\limits_{x \to 0}(\ln x+\ln \frac{1}{x} )$$ Since $\ln x+\ln \frac{1}{x} =\ln x\frac{1}{x}=\ln 1=0$, then $$\lim\limits_{x \to 0}f(x)+ \lim\limits_{x \to 0}g(x)= \lim\limits_{x \to 0}(\ln x+\ln \frac{1}{x} )=\lim\limits_{x \to 0}(\ln x \frac{1}{x} )=0.$$ Thus, the limit of the sum exists, even if the individual limits do not.