## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to4}f(x)=7$
We have $\lim\limits_{x\to4}(4x-9)=4\times4-9=7$ $\lim\limits_{x\to4}(x^2-4x+7)=4^2-4\times4+7=7$ Therefore, $\lim\limits_{x\to4}(4x-9)=\lim\limits_{x\to4}(x^2-4x+7)=7$ But, $4x-9\leq f(x)\leq x^2-4x+7$ for $x\geq0$ So, applying the squeeze theorem, we have $\lim\limits_{x\to4}f(x)=7$