## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to a}f(x)=0$$
For all x, we have $$|f(x)|\le g(x)$$ which means $$-g(x)\le f(x)\le g(x)\hspace{1cm}(1)$$ Also, we have $\lim\limits_{x\to a}-g(x)=-\lim\limits_{x\to a}g(x)=-0=0=\lim\limits_{x\to a}g(x)\hspace{1cm}(2)$ Therefore, from (1) and (2), according to Squeeze Theorem, we can conclude that $$\lim\limits_{x\to a}f(x)=0$$