## Thomas' Calculus 13th Edition

a. $5$ b. $5$
a. Given $\lim\limits_{x \to 2}\frac{f(x)-5}{x-2}=3$, we have $\lim\limits_{x \to 2}f(x)=\lim\limits_{x \to 2}3(x-2)+5=5$ b. Given $\lim\limits_{x \to 2}\frac{f(x)-5}{x-2}=4$, we have $\lim\limits_{x \to 2}f(x)=\lim\limits_{x \to 2}4(x-2)+5=5$