#### Answer

(a) The slope of the secant line $PQ$ is $$m_{PQ}=\frac{f(x)-f(3)}{x-3}$$
(b) The slope of the tangent line at $P$ is $$m=\lim\limits_{x\to3}\frac{f(x)-f(3)}{x-3}$$

#### Work Step by Step

(a) The secant line is the way to intuitively come up with the way to calculate the slope of the tangent line at any point of a graph.
Normally, if we already know 2 points of a line, for example point $A(a, b)$ and point $B(c, d)$ of line $(d)$, then the slope of line $(d)$ is calculated as follows: $$m_{(d)}=\frac{d-b}{c-a}$$
Therefore, in this case, since we know point $P (3, f(3))$ and $Q (x, f(x))$, the slope of $PQ$ would be $$m_{PQ}=\frac{f(x)-f(3)}{x-3}$$
(b) As we move $Q$ closer and closer to $P$ along the curve $y=f(x)$ by getting $x$ closer to $3$, eventually we would get $Q$ in the place of $P$, then we would get a line $(d)$ which does not cut across the curve $y=f(x)$ anymore.
Then $(d)$ is the tangent line of the curve $y=f(x)$ at $P$.
In the process, the slope $m_{PQ}$ also changes towards a number $m$, which is the slope of the tangent line $(d)$.
Then $m$ is actually the limit of $m_{PQ}$ as $x$ approaches $3$.
In other words, $$m=\lim\limits_{x\to3}\frac{f(x)-f(3)}{x-3}$$