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

$g(x)=\frac{df(a)}{dx}(x-a)+f(a)$
Given that $g(x)=m(x-a)+c$ condition 1: $E(a)=0$ and condition 2: $\lim\limits_{x \to a}\frac{E(x)}{x-a}=0$ now $E(x)=f(x)-g(x)=f(x)-(m(x-a)+c)=(f(x)-m(x-a)-c)$ on applying condition 1 to the above equation: $E(a)=f(a)-m(a-a)-c=0$ $c=f(a)$ on applying condition 2 to the above equation: $\lim\limits_{x \to a}\frac{E(x)}{x-a}=0$ $\lim\limits_{x \to a}\frac{(f(x)-m(x-a)-c)}{x-a}=0$ $\lim\limits_{x \to a}\frac{(f(x)-c)}{x-a}-m=0$ $\lim\limits_{x \to a}\frac{(f(x)-f(a))}{x-a}-m=0$ $\frac{df(a)}{dx}-m=0$ $\frac{df(a)}{dx}=m$ on substituting the value of g(x): $g(x)=\frac{df(a)}{dx}(x-a)+f(a)$