## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to0}g(x)=2$$
- Calculate $\lim_{x\to0}(2-x^2)$ and $\lim_{x\to0}(2\cos x)$ $\lim_{x\to0}(2-x^2)=2-0^2=2$ $\lim_{x\to0}(2\cos x)=2\times\cos0=2\times1=2$ So, $\lim_{x\to0}(2-x^2)=\lim_{x\to0}(2\cos x)=2$ - Yet $2-x^2\le g(x)\le 2\cos x$ for all $x$ Therefore, applying the Sandwich Theorem, we conclude that $$\lim_{x\to0}g(x)=2$$