Answer
(i) Using the first definition
$m = \lim\limits_{x \to 1} = \frac{f(x) - f(1)}{x-1} = 2$
(ii) Using the second definition
$m = \lim\limits_{h \to 0} = \frac{f(1+h) - f(1)}{h} = 2$
(b) Tangent line
$y = 2x+1$
Work Step by Step
(i) Using the first definition
$m = \lim\limits_{x \to 1} \frac{f(x) - f(1)}{x-1}$
$m = \lim\limits_{x \to 1} \frac{(4x-x^{2}) - (4(1)-1^{2})}{x-1}$
$m = \lim\limits_{x \to 1} \frac{-x^{2} +4x - 3}{x-1}$
$m = \lim\limits_{x \to 1} \frac{(x-1)(3-x)}{x-1}$
Cancel out $(x-1)$
$ m = \lim\limits_{x \to 1} (3 -x)$
$ m = (3 -1)$
$ m = 2$
(ii) Using the second definition
$m = \lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$
$m = \lim\limits_{h \to 0} \frac{[4(1+h) - (1+h)^{2}] - [4(1) - 1^{2}]}{h}$
$m = \lim\limits_{h \to 0} \frac{2h - h^{2}}{h}$
Simplify:
$m = \lim\limits_{h \to 0} \frac{h(2- h)}{h}$
$m = \lim\limits_{h \to 0} 2 - h$
$m = \lim\limits_{h \to 0} 2 - 0$
$m = \lim\limits_{h \to 0} 2= 2$
(b) Tangent Line
$ m = \frac{y-3}{x-1} $
$m = 2$
$2 = \frac{y-3}{x-1} $
$2(x-1) = \frac{y-3}{x-1} (x-1)$
$2x - 2 = y-3$
$ y = 2x -2+3$
$ y = 2x + 1$
