Answer
$m=-\dfrac{1}{5}$
Work Step by Step
$f(x)=\dfrac{5}{x+2},$ at $(3,1)$
The slope of the tangent line at the point $P(a,f(a))$ is given by $m=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=3$
Find $f(a+h)$ by substituting $x$ by $3+h$ in $f(x)$ and simplifying:
$f(3+h)=\dfrac{5}{3+h+2}=\dfrac{5}{5+h}$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=1$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{\dfrac{5}{5+h}-1}{h}=\lim_{h\to0}\dfrac{\dfrac{5-(5+h)}{5+h}}{h}=...$
$...=\lim_{h\to0}\dfrac{\dfrac{-h}{5+h}}{h}=\lim_{h\to0}\dfrac{-h}{h(5+h)}=\lim_{h\to0}\dfrac{-1}{5+h}=...$
$...=\dfrac{-1}{5+0}=-\dfrac{1}{5}$
