Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to1}f(x)=1$$
1) Since $2x-1\leq f(x)\leq x^2$ for $0\lt x\lt3$ and $1\in(0,3)$ 2) Consider $\lim\limits_{x\to1}(2x-1)=2\times 1-1=1$ $\lim\limits_{x\to1}x^2=1^2=1$ That means $\lim\limits_{x\to1}(2x-1)=\lim\limits_{x\to1}x^2=1$ 3) From 1) and 2), according to the Squeeze Theorem, we can conclude that $$\lim\limits_{x\to1}f(x)=1$$