## Calculus (3rd Edition)

$$\lim _{x \rightarrow \pi / 2} f(x)=1$$
We are given the functions $$u(x)=1+\left|x-\frac{\pi}{2}\right|\quad \text { and }\quad l(x)=\sin x$$ From the following figure, we have the limits$$\lim _{x \rightarrow \pi / 2} l(x)=\lim _{x \rightarrow \pi / 2} u(x)=1$$ Then by using the squeeze theorem $$l(x) \leq f(x) \leq u(x)$$ Then $$\lim _{x \rightarrow \pi / 2} f(x)=1$$ In other words, any function $f(x)$ will be "squeezed" toward $1$ at $x$ tends toward $\pi/2$.