This is a rather sophisticated problem, but is worth working through because it brings
together a number of the key subtleties of limits and continuity, and is very instructive.
We have
f(x)=x− 3
x^2 − 4 x+ 3=x− 3
(x− 3 )(x− 1 )This has two discontinuities, atx=1andx=3. Atx=3 the function is indeterminate,
but otherwise we can cancel thex−3 and write
f(x)=1
x− 1providedx
= 3Atx=1 the discontinuity is infinite, of the same kind as that of 1/xat x=0(cf:
Problem 14.4),exceptthat the pointx=3 must be omitted, see Figure 14.7.
y0 1 3 x
− 1Figure 14.7The functionf(x)=
(x− 3 )
(x− 3 )(x− 1 )
.
Whilef(x)does not exist atx=3, the limit of the function does:lim
x→ 3
f(x)=lim
x→ 3x− 3
(x− 3 )(x− 1 )=lim
x→ 31
(x− 1 )
=^12So we can ‘plug the gap’ and form a continuous function fromf(x)forx>2 by defining
a new functiong(x)that is equal tof(x)forx>2,x
=3, but is equal to^12 atx=3:
g(x)=f(x) x > 2 ,x
= 3
=^12 atx= 3