Answer:
Given function is:
[tex]f(x)=\frac{3}{x}\text{ at x=3}[/tex]
To find the slope of any tanget line using limit definition.
we get,
Slope m is
[tex]m=\lim _{h\to0}\frac{f(3+h)-f(3)}{h}[/tex][tex]=\lim _{h\to0}\frac{\frac{3}{3+h}-1}{h}[/tex][tex]=\lim _{h\to0}\frac{3-3-h}{h(3+h)}[/tex][tex]=\lim _{h\to0}\frac{-h}{h(3+h)}[/tex][tex]=\lim _{h\to0}\frac{-1}{3+h}[/tex][tex]=-\frac{1}{3}[/tex]
Slope is -1/3
f(x) at x=3 is,
Put x=3 in f(x), we get
[tex]\begin{gathered} f(3)=\frac{3}{3}=1 \\ f(3)=1 \end{gathered}[/tex]
we know that, the equation of line with slope m and point (x1,y1) is given is
[tex]y-y1=m(x-x1)[/tex]
Here m=-1/3 and (x1,y1)=(3,1)
Substitute we get,
[tex]y-1=-\frac{1}{3}(x-3)[/tex][tex]y-1=-\frac{1}{3}x+1[/tex][tex]y=-\frac{1}{3}x+2[/tex]
The required equation of a tangent line at x=3 is y=-1/3 x+2.
